LIGHT - REFLECTION & REFRACTION Class 10 Physics | Complete One Shot in Pure English

Hello class 10th students. What if I ask you to stand in front of the plane mirror? What you will observe? Your answer will be very simple. So image will be obtained. But what you are not going to obtain that is the kind of fear you are having for your board examination because physics seems tough for you. So don't worry students physics walla that is the most trusted education platform bought you a 10th board booster For you. That is a free crash course. So in this very first chapter that is light reflection and refraction. I will

be covering every single corner that is reflection of light its laws, reflection through plane mirror, reflection through spherical mirrors, refraction of light, refraction through spherical lenses, power of a lens and power of combination. These are the major topics which has been asked in the previous year questions. So I will be covering All the concept along with the numericals as well as the previous year questions which are actually required for you. So me your physics captain Vikas Indraal will be providing you all the lectures of physics of class 10th and to watch all the lectures you

have to go to the PW foundation English channel and to get class notes DPP practice sheet what you have to do you have to click on the link given in the description box which will redirect you To the PW application you have to enroll for the free batches apart from that you will be getting class notes DPP practice sheets so that you can ensure your preparation is 100% % perfect before the board examination and this is my guarantee and I am count on my words student you have to put complete faith what I'm going to

teach you have to observe everything you have to learn all the concept practice everything and you will be an achiever so let's start so Today we will be taking your first lecture of physics that is light comprises of two different categories that is reflection of light and refraction of light you know what the manual students are getting confused sir there are too many numericals are there sir there are too many rules there are formation of images I have a plan for you so that you can make it more easier I'll try to explain each and

every single thing according to what I just Said to you in that manner it will be quite convenient for you to understand the concept so what exactly I have so the topics that we are going to cover that will in a proper sequence that is reflection of light. Then we will discuss about spherical mirrors. Then sign conventions then rules and formation of images by spherical mirrors. You need to put more concentration in that particular part and refraction of light then spherical Lenses. Rules and formation for the spherical of lenses and power of the lens. This

is how in eight different parts we will we will be covering the entire chapter. So let's begin. So the introduction says what is light? The very first thing or a one marker question appears from that like what is light? So we need to understand the light is a form of energy. It is nothing but a form of energy which helps us to see things around us. See without light You won't be able to see anything. It will be a complete dark. So in the presence of light we are able to see. So it is a

form of energy through which you are able to see or it gives a sensation of vision. So that's how you do that. And the speed of light in vacuum that is 3 * 10 ^ 8 m/ second. So that is the speed of light which we are going to use at various places. So you need to remember this value. Sometimes people you know the students are asking so what Is the speed of light in air so it is approximately 3 * 10 ^ 8 because there is there is a presence of particles or a medium

that is air medium so speed gets a bit lower side as compared to the vacuum but approximately we consider it as 3 * 10 ^ 8 m/s how the light travels so the light travels in a straight line so it always travel in a straight line so whenever we are going to draw the ray of light or the beam it will always travel in a Straight path. Apart from that from where the light is going to come. So we we know about the sun bulb light. So these are called sources. So the source of light

is nothing but an object from which light is given out maybe natural or maybe man-made as I gave you example of sun. So that is a natural source of light. bulb, light, these are the man-made sources of light. Apart from that, when we talk about the images, there are two kinds of images. What is That? The very first is real and inverted. We need to understand what is real and inverted. So, it's quite easy. If the object is like that, the image will be appearing upside down. So, that is called inverted image. That is called

inverted image. If the object is like that the upright the image will be forming upside down that is called the inverted image. What about the real? So in a real category I would be writing which can be obtained On screen which can be obtained on screen. So real images are those images which can be obtained on screen. The example is the movie theater. The movie which you generally watch in a theater you see a projector which is projecting on a particular screen that is called the real image right the projector you must have seen in

your school then they must be showing some presentation or some video. So you are required a particular Screen. Such images are called real images. Whereas there are other kind of image that is virtual and erect. Virtual image which cannot be obtained on screen. Which cannot be obtained on screen. Like example, let me tell you when you are standing in front of the plane mirror, do you require any screen to see your image? No, never. You do not require any screen to see your image. What you see? You see yourself inside the mirror. So you do

not require any screen. So in that case, such particular images are called virtual images. What about the erect property? So when you are standing in front of the plane mirror, you are standing like that. the image will be exactly like that. So the in this is called inversion of the image. This is called erect image. So that is the difference between the real and inverted and virtual and erect. Sometime the question you know the student might ask that can a virtual image can be inverted? Generally no no virtual image will be erect and real image

will be inverted. The next thing is what is a medium? Because the light is traveling through various mediums. So what is a medium? So medium is a substance through which light propagates through which the light travels. Reflection of light. So that is one of the phenomena which we generally You know observe in our daily life. So reflection is nothing but it is the bouncing back of light in the same medium on striking the surface of any object. You must have seen the plane mirror. Let's take an example for the plane mirror. So that is a

plane mirror. Consider it as plane mirror. I'm doing the polishing on the right side. So now it became a plane mirror. So if here we have a medium. So let's suppose this is the medium and light is Striking on the plane mirror. It will reflect back in the same medium. This is what the reflection of light is. The bouncing back of light in the same medium is called reflection of light on striking on any surface be it plane mirror or spherical mirror or any other surface. So this is what the reflection of light is. What

kind of different reflections are there? So there are two kinds of reflection. One is regular reflection and second is irregular or Diffused reflection. Irregular reflection can also be called as diffused reflection. So regular reflection is what? So regular reflection says when the reflecting surface is smooth like plane mirror and well polished again like plane mirror the parallel rays falling on it are reflected parall to one another. If I draw the another ray of light parallel to this ray after reflection it will be reflecting back parall to the reflected Ray. So you can see this is

what the you know regular reflection is. So understand the concept when the rays of light is striking on any surface after reflection it is going again parallel in the same medium that is called the regular reflection. Whereas the irregular reflection will be exactly opposite to that. Like if the rays of light are coming parall to each other striking on surface going in the various diffused direction that is called the Irregular or it can be also called as it can also be called as diffused reflection. See we are able to see different objects like your wall,

ceiling, floor. They are not perfectly reflecting material but we are able to see them because they are actually doing the irregular or diffused reflection. The reason is there is no polish no shiny surface. So when the light is striking on it, it is having some sort of diffused surface. So the Light diffused in the various direction and we are able to see the object. So such reflections are called diffused reflection. Let's take an example. So we are having the different categories over here of the regular reflection and irregular reflection. So let's see the rays of

light are coming parall to each other and I'm drawing the normal. This is what the normal par perpendicular to the surface is a normal. So if you observe All the light rays are going parallel after reflection whereas in the diffused reflection if you see the light is going in the various direction like that like that. So this is the difference between regular and irregular reflection. Moving ahead. So what are the laws of reflection? one of the favorite questions or you know the analysical based questions which is appearing in the previous exam years based on such

diagram which is the Explanation of law of reflection. So the according to the laws of reflection that we have two different kind of laws. The very first law says the angle of incidence is equal to the angle of reflection. Amazingly what are the notations? So the light which is actually incidenting we need to understand this. If the light if the ray of light is incidenting on the plane mirror that is called the incident ray. Whereas the the light which is going Back that is called the reflected ray and the angle formed between incident ray and

normal that is called the angle of incidence. And the angle formed between reflected ray and normal is called angle of reflection. And this is my normal. Normal is nothing but perpendicular and imaginary perpendicular to any surface at any point. So according to the law of reflection, it says the angle of incidence will be equal to the angle of Reflection. They will be equal to each other always and always. Be it for the regular reflection, be it for the irregular reflection, be it for the spherical mirrors, it is applicable everywhere. The second point is the incident

ray, reflected ray, normal all lie on the same plane at the point of incident. So this is the point of incident. So they all lie on the same plane at the point of incidence. Most important question, two marker. Sometimes in an analytical question, four marker. So just memorize it. Law of reflection will be applicable everywhere. It is a law not a rule for anything. So law will be applicable everywhere. Moving ahead. So the reflection of light in this category like what are the exact characteristics which we generally see in the plane mirror. When we see

the images in the plane mirror what we observe the very first thing that we observed which is The image of a real object is always virtual. So the images formed by the plane mirror will be virtual and it cannot be obtained on screen. It cannot be obtained on screen and it will be a virtual image. I already gave you the insight about the virtual and direct image and this is what we are going to get in the plane mirror. So this is the very first point that we have covered. The next point the image formed

in the plane mirror is always erect. So if You're standing upside right, it will give you the same image. Your head will be up as the object. So it will give you always an erect image. Like the upside of the image is the upside of the object. This is how you get the image in the plane mirror. The size of the image in the plane mirror is always the same as the size of the image. Which means whatever your height is, whatever the size of the object is, the image will be also of the Exact

same size when you are obtaining the images in the plane mirror. That is point number three. So one, it gives a virtual image. Two, it gives an erect image. Three, the size of the image is equal to the size of the object. So these are the three points which we have seen so far. The other two points are the image formed in the plane mirror is as far behind the mirror. That means if this is your plane mirror, if the object is here, the image will be exactly on The other side at the same distance

from the plane mirror. If you need to observe this, so this is how you see yourself. Like for an example, if this is my plane mirror and I'm drawing an object here, you will be getting the image exactly opposite. The distance between the plane mirror and the object and the distance between the plane mirror and the image will be equal to each other. If they are standing d distance apart, the distance of the object from the plane mirror will Be equal to the distance of image from the plane mirror will be equal to each other.

Point number five, the image formed in the plane mirror is literally inverted. You must have noticed that when you're standing in front of the plane mirror and observing your image, if you are taking your right hand up, it seems like in the image you are taking your left hand up. So that is called the lateral inversion. If you see the image of the boy, he is, you know, already Taking his right hand up. So in the image, it appears like he's raising his left hand in the upward direction. So that is called the lateral inversion.

It is not inverted image. It is called lateral inversion. So these are the major five characteristics what we usually obtain when the image is formed in the plane mirror. The first was virtual, second erect, same size as the object, same distance from the plane mirror of the image and the last is Lateral inversion. This is how you get the five characteristics of the formation of images in the plane mirror. Talking about spherical mirrors. Now from here you know the the thing is going to be complicated for you. We'll keep it sorted. You just only need

to understand the concept. Mind my words. Whatever I'm saying I'll make it quite easier for you and things will be sorted for you. Just listen carefully what I'm going to say in next half an hour. So Spherical mirrors just imagine you have a complete sphere right in front of you like that whatever is given in the diagram you have a complete hollow sphere and you just given a small cut and you get a small bowl kind of thing like that that is the spherical mirror. So I have taken a sphere which must be having a

center in the middle of the sphere. I have taken a part out of it I have got an this kind of image or the figure I Can say. Now just listen to this carefully. There are two different diagrams which you can see right here. The very first one is this. Second one is this. Here the polished surface you can see on the right side. Whereas the polish you can see on the left side. So this is how we get the two different kind of spherical mirrors. But what are the names of the spherical mirrors? So

one is the concave mirror, other one is Convex mirror. But we need to identify which one is concave and which one is convex. So talking about the spherical mirrors, I'm I'm giving you a very easy method to remember which one is concave mirror and which one is convex mirror. This is my hand. Consider it as a spherical mirror. If I'm going to polish the outer surface and the reflection taken from the cave side, I'm repeating if the polish is done on the other side and the Reflection is taken from the cave side, then it is

called a concave mirror. on the opposite head. If I'm going to polish the cave side and the pol and the reflection taken from the burst outward part that is called the convex mirror. This is how you can remember which one is concave which one is convex. Go to the previous image. Now you can see the reflection is going to take place from the cave part. So this particular mirror is nothing but concave Mirror whereas the cave part is polished and the reflection will will be you know taken from the bulged outward part that is called

the convex mirror. I hope you understand what exactly I'm going to say. So if you understand this we need to understand what are the major points or terminologies we need to remember to understand the concept of spherical mirrors and the formation of images by the spherical mirrors. So consider we Have these two mirrors right in front of me. One is concave mirror other is convex mirror. There are certain points which I'm going to mention. The very first one, this point that is called pole represented by P that is called pole represented by P that is

the midpoint of the spherical mirror whatever you are going to take be it convex or concave. So the middle point of the spherical mirror that is called pole. We have the Center of curvature. So what does that mean? Center of curvature is the center of that sphere of which this mirror is a part. So you have taken a sphere you have taken out this spherical mirror that sphere was having a center that is a imaginary center that we are considering of which it was a part that is called the center of curvature for both the

mirrors represented by capital C. Now if we have an imaginary line passing through pole and center of Curvature. I'm having an imaginary line passing through pole and center of curvature. That is called a principal axis. So three things we have covered so far. One is pole. Second is center of curvature. And the principal axis. Principal axis is an imaginary line passing through pole and center of curvature. The distance between see all the distances will be measured from pole. That is one of the rule. The distance between pole and center of Curvature that is called radius

of curvature represented by capital R represented by capital R. It is the distance between pole and center of curvature for both the mirrors. Just make a note of it. I'm going a step ahead to make you understand like what is the definition. So for pole the center of the spherical surface of the mirror this is what I told the center of the spherical surface of the mirror center of curvature is the of which the Mirror is a part the center of the spherical shell of which mirror is a part it lies outside the surface of

the mirror you must have observed the mirror was placed like that and the center was outside the mirror that is called the center of curvature represented by capital C principal axis is an imaginary straight line join ing the pole of the mirror and center of curvature. Principal focus you guys need to focus. Principal focus Is the point on the principal axis of the mirror such that rays incident on the mirror parall to the principal axis. If the rays of light coming parall to the principal axis after reflection they meet or appear to meet at this

point. So we need to understand what does that mean like meet or appear to meet in few minutes. Radius of curvature will be the distance between the pole and the center of curvature. This is what the radius of curvature is talking about the focal Length. See generally the focal length appears between exact between the pole and the center of curvature. But the distance between pole and focus will be called as focal length. The focal length is the distance between pole and principal focus of the mirror. That is called the focal length. And we have the

aperture. Aperture gives the size of the mirror. It gives the size of the mirror. So I have taken like this much size of the concave mirror for example. So this Is called moving ahead talking about the principal axis. How we need to define or how we exactly are going to define the principal axis. So for that I need to make few points that is pole that is center of curvature I need to draw principal axis which is the imaginary line passing through the pole and center of curvature as I mentioned for example if the rays

of light coming parall to the principal axis if the rays of light coming parall to the principal Axis after reflection what is going to be happen. Listen to this. If you are having a plane mirror or a plain wall, if you are throwing a ball right in front of the plain surface, after striking it will be coming back straight. You know that. What if the surface is like that? So if the ball is going to strike at such surface, it is going to come down and at the bottom it is going to come up. Same

is going to be happen with the light rays. So after reflection this ray of light will be coming down and this ray of light will be going like that and this point that we have obtained that is called focus and the distance between focus and pole will be considered as focal length represented by small f represented by small f. So we need to make this difference very clear. Principle focus when we are defining the point you need to write it as capital F when you are measuring the distance Between the pole and focal length it

will be a small F that is focal length got it so this is principal focus this is focal length talking about the convex mirror so we have pole the curve is on the right side so the curvature or the center of curvature will be also on the right hand side so if I draw the principal axis and the rays of light coming parallel to the principal axis. Now that is an interesting case As I gave you an example you are throwing a ball in front of the plane mirror or the concave mirror. Let's do the

same activity with the convex mirror. So if the surface is like that you will observe that the rays will be diverging like that. The rays are diverging after striking on the mirror. They will diverge in that direction and it has been said that they will appear to meet. They are Actually not going to meet you know for by the end of the time they're not going to meet but they will appear to meet at this point and this point is called principal focus for the convex mirror and the distance between pole and principal focus will

be again represented by small f the most interesting and you know most frequently asked question in the board examination. Moving ahead like express the relation between focal length and radius of Curvature of by of spherical mirror. They're asking about the relation like what is the difference between radius of curvature and focal length. So the relation is quite easy. The radius of curvature the radius of curvature is twice of focal length or focal length is equal to half of the radius of curvature. That is the relation between focal length and radius of curvature. I hope you

got it. Now rules for the formation of image is very very very very important because the numericals which you need to solve. It's not only by mathematics. You need to understand how the rays are coming where the image will form. What kind of nature it is going to give. So before forming the images you need to understand the rules. It is exactly like when you are driving a car you need to understand the traffic rules. This is how we are going to Understand the rules for the formation of images. The very first rule says the

incident ray parall to the principal axis actually passes or appear to pass through the focus. Yes. Yes. Yes. In the previous slide just we have seen when the rays of light was princing parall to the principal axis it was actually meeting at focus when we have seen in the concave mirror and it was appearing to pass through focus when we are looking into the Convex category. So that is the very first rule. Second rule, incident ray through the center of curvature falls normally and reflected back along the same path. when the ray of light when

it is incidenting or appear to you know passing through the center of curvature like for example if I draw this so that is the principal axis and I'm mentioning this is my center of curvature this is pole and if a ray of light and if a ray of light passing Through the center of curvature. It will be coming back along the same path. No changes required because it was already passing through the center of that sphere of which the mirror was part. Same is going to be happen same is going to be happen with the

convex mirror. So that is the convex mirror. This is pole. This is center of curvature from principal axis. This is the principal axis And the rays of light coming parall to principal axis. The rays of light passing through center of curvature it is exactly not going to do that. So we need to appear to pass through the center of curvature. So it seems like it is passing through the center of curvature. So after reflection it will be following the same path. It will be following the same path. So whatever light is going through the center

of curvature it will follow the same path After reflection. Point number three. So look at this. The same thing is given like rays of light coming parallel passing through the focus and for the center of curvature it will be passing through the center of curvature only. Point number three or rule number three. If a ray of light incident through the focus is reflected parall to the principal axis, it is exactly opposite to the rule number one where the rays of light was Coming parall to the principal axis passing through focus on the opposite and if

the rays of light passing through the focus it will be going back parall to the principal axis. That is our rule number three. So look at this image. If the rays of light passing through the focus, it will be going back parall to the principal axis. In the convex, if the ray of light appears to pass through focus, it will be going back parall to the principal axis. That is rule number Three. And the last rule says if the ray of light incidenting towards the pole by making some angle is reflected back on the same

side by making the same angle. For this we need to understand this pole. On which mirror? Concave mirror. The ray of light will be going back by forming the same angle. Which means again angle I equal to angle R which is happening in the previous cases also. But here we can clearly see that it is it will be making same angle. Same as With the convex. So if the ray of light incidenting on pole it will be forming the same angle and that that will be equal to each other angle I equal to angle R.

So these are the four rules. I'm again going to repeat what are the four rules which we have seen just now. The very first rule if the ray of light passing through principal axis it will be passing through focus. If the ray of light passing through center of curvature it will be reflecting back Along the same path. If the ray of light passing through focus it will be going parall to the principal axis and the rule number four if the lay ray of light incidenting on pole it will reflect back along the same medium by

making or by forming the same angle. Moving ahead to the formation of images. In concave mirror, we have six different cases. Whereas in convex you will be surprised that we have only two cases that eight cases we need to understand how this is Going to be happen. Rule number or the formation number one which says when the object is placed at infinity. So we need to understand where the object is being placed. So we are considering this is my spherical mirror and the object according to the size the very very far away that will be

considered as infinity and the light rays which are coming through the infinity will be almost parall to the principal axis which you can see right away right here. So these Are the light rays which are coming parall to the principal axis. Such light rays after reflection it will be passing through focus. That is the rule number one that we know. If the rays of light coming parall to the principal axis they will meet at focus. So that is also applicable for the formation of image when the object is placed at infinity. Now what kind of

other things that we are going to get? So the object is at Infinity image that we obtained at focus. If we talk about the size, it seems like points sized point sized and nature of image will be real and inverted. Remember when I explained about the two types of images, it is one of them. So the light rays will cross each other. So we are required a paper or a screen to get the image and you will be getting a point Image over here. So this is why it is a real and inverted image. That

was formation number one. Now I'm going to shift the object a bit towards the mirror. And this is what it looks like when the object is placed beyond center of curvature. Initially it was very far. Now I have taken it to a considerable distance but it is just beyond center of curvature. And we will be following the rules. According to the rules, if the rays of light going parall to the Principal axis, it will be passing through focus. Rule number one. That's what we have seen. Rule number two. So if the rays of light passing

through focus in the concave mirror after reflection it will be going parall to the principal axis. Yes, these are the two you know vice versa rules that we have seen. And now I can see the image or the reflected rays are meeting at this particular point and at the same point the image is going to be formed. Here you can see the image is like that. So if the name of the object is like for example AB the image will be a - b dash. But sir why this object is upside down because the nature

of image is real and inverted. So again object is beyond center of curvature beyond center of curvature. So the image obtained between curvature center of curvature and focus. Talking about its size, it will be comparatively smaller. It will be comparatively smaller. So I can write it as diminished for the smaller. I can use this term diminished and nature will be real and inverted. Now again I need a screen to get this image. So that was the formation number two. Now I'm again going to take my object bit towards the center of curvature and I have

got the point that is the center of curvature itself. So if the object is at center of curvature follow the rules that I have explained. So if the rays of light going parall to The principal axis the reflected ray will be passing through focus. So that is the incident ray after reflection it is passing through focus. Perfect. And again the second rule if the ray of light passing through focus it will be going parall to the principal axis and I have got the image beautiful image and what I have observed I have seen that the

image is forming on center of curvature itself. Yes because there was a sphere of which the mirror was a part. If you are going to take the object at the center the image will be exactly at the center. So you have obtained the image at the center of curvature itself. So that is a very very interesting case. If the object is at center of curvature, image will be also at center of curvature. Talking about size same as object advantage of being placed at the center of curvature. uh nature that is real and inverted Because again

you required a screen to obtain the image. So that was formation number three. Moving ahead to formation number four. I have just taken my object at this place naming it as AB following the rules which says if the ray of light going parall to the principal axis it will be passing through focus and if a ray of light passing through focus after reflection it will be going parall to the principal axis and I have Seen that the image formed beyond center of curvature at this point and the name is a dash bdash. Interestingly, if you

have noticed in the case number two where the im object was placed beyond center of curvature, image was between center of curvature and focus. Now when I'm placing the object between center of curvature and focus, the image obtained beyond center of curvature. So if I'm going to place object between C and F The image will be formed beyond center of curvature. Talking about its size. If you have noticed initially when the object was far it was a point image beyond C it was diminished. At center of curvature it became same. Now the size of image

is going to increase and it will be a enlarged image in this case and talking about the nature nature will be real and inverted once again. Now that was formation number four Formation number five. Now I have kept the object at focus. So that's my object. Now the rays of light passing through focus. Now these are the two light rays passing through focus. And if the rays of light passing through focus it goes parallel to the principal axis after reflection. Yes. Where they are going to meet? Nowhere. Because these are the parallel rays. Parallel rays

never intersect each other. They meet at infinity. Yes, you Are right. they meet at infinity. So I'm going to write if the object is at focus the image will be at infinity size of the nature that will be highly highly enlarged. highly enlarged and nature to get the image of these light rays you need real and inverted image. So on screen when you will be obtaining the image of this you will be getting a real and inverted image. Now these are the major five cases that we have seen so far at Infinity beyond center of

curvature at center of curvature between center of curvature and focus. at focus. Now the twist begins. Now I'm going to place my object between principle focus and pole and I will be obtaining something different. So let's do it. So I have taken this object naming it as AB following the rules which says if the ray of light going parall to the principal axis after reflection it will be passing through focus and if the ray Of light passing through the center of curvature it will be following the same path. So if you see they are neither

parallel nor meeting each other. They're almost like you know some sort of diversion. So they are never going to meet but they will appear to meet somewhere at the back which gives the formation of image as virtual and erect image. So here we will be obtaining the image A-B dash. So when you are putting your object between Pole and focus the image will be behind the mirror. Behind the mirror as plane mirror nature will be different. If you obtain the size of image that will be enlarged you will be getting a enlarged image. And talking

about the nature that will be virtual and direct this time virtual and erect. So there is only only and only one case in a concave mirror where you will be obtaining virtual and erect image and this is when when you Are putting your object between focus and pole. Got it? So these are the six formation of images by concave mirror. Before that now there is a complete chart you can see. Have a look. Interesting to learn object at infinity image at focus highly diminished or point size real and inverted. When you put beyond center of

curvature image will be between f and c diminished Image will be real and inverted. If the object is at center of curvature, it will be forming at center of curvature. Same size, real and inverted. Between focus and center of curvature, image will be beyond center of curvature. Now it is going to be enlarged. Nature will remain real and inverted. Now the object at focus it will be forming at infinity. Highly highly enlarged nature will be real and inverted. At the last case when the Object is between pole and focus it will be behind the mirror

enlarged. Now the nature is virtual and erect. Moving ahead to the convex mirror. So there are two cases that we need to understand. First when the object is at infinity what is going to be happen? So let's consider the light rays are coming from infinity from outside somewhere like this. So what is going to be happen if the ray of light incidenting on pole it will be Reflecting back by forming the same angle. If the ray of light appearing to passing through pole it will be reflecting back parall to the principal axis. If the ray

of light appear to pass through center of curvature, it will be reflecting back. And all this what we have observed where at focus where we have observed when the object is at infinity image will be at focus. Talking about the size that will be highly Diminished or small, highly diminished and nature here will be virtual and erect because it is forming inside or behind the mirror. So nature will be virtual and erect. Second and last case if the object is placed anywhere anywhere between infinity and pole what is going to be happen so let's consider

this is your object AB naming it as AB Following the rules if the ray of light going parall to the principal axis passing through focus or appear to pass through focus as you can see it is passing through focus appear to pass. Other one if the ray of light appear to pass through center of curvature, it will be following back. Now here you can see the ray of light appear to pass through center of curvature. What I have seen these appeared line incidenting at a point. Now this is the image A dash B dash and

I will conclude like the object will be anywhere anywhere between infinity and pole image will be between pole and focus always it will be between pole and focus. talking about size always be diminished always be diminished and nature that will always be virtual and erect. So these are the two cases. Let me tell you the uh uses of concave mirror and convex mirror. So the very first thing uses of concave mirror. So testing of teeth by concave mirror. So generally dentists use it to see the larger image by using concave mirror because when the object

is placed between pole and focus it will be forming virtual image enlarged image. So this is how the teeth or the tooth cavity would be you know enlarged. It will be visible to the Dentist and that can be cured. Second, if you have seen the burning mirror where you know the concave kind of surface being used in a torch light in a solar cooker. You must have familiar with the solar cooker in third four from third fourth grade and the headlights of the car. Talking about convex mirror. So it it has been used as a

so let's talk about the uses of convex mirror. So the uses of convex mirror you must have seen the rear view mirror in various cars two Wheelers where generally it has been written like objects may appear while than they actually are. So we it gives the smaller images and it gives protection to us so that we can see the vehicles are coming from back. So we can see that and it gives the wider view because it generally gives the diminished images. Similarly nearby your colony or maybe at the sharp turns or the U-turns you must

have seen this kind of mirror which is actually a convex Mirror so that you can see the small images of the objects which is coming from the other end and you would be able to identify and you will be feel safe so that how to take turns. So this kind of mirror will be very useful. Moving ahead to the sign convention. What are the sign conventions required for solving the numericals or to understand the value? So remember few points are there. So sign conventions for measuring distance from the convex and concave. First point the very

first point that the all the distances that we have done so far all the distances will be measured from pole. All the distances like whatever distance you need to take radius of curvature, focal focal length, object distance, image distance that will be considered from pole only. Second, the incident ray is taken from left to right. So wherever you are going to put your spherical mirror concave or convex, the incident ray that we we need To take from the left side only be it concave be it convex. So incident ray will be coming from left to

right. That is point number two. Point number three distances measured in the same direction as that of the incident ray can be taken as positive. If we are taking this direction as positive and that is the direction of incident ray. So we will be taking this direction as positive from the pole. Whereas opposite to that will Be considered as negative. That was point number three and four. So if you are taking this direction consider it as positive direction. If you are taking this direction in that particular way that will be considered as negative direction. Point

number five distances measuring upward from the principal axis will be considered as positive. Like for suppose this is my principal axis. I have taken an object upright AB and this will be considered as positive because This is in the in the positive y direction. Whereas if I get a real inverted image that will be considered as negative because it is in the negative y-axis. So this is how we need to learn and another method to understand this concept I'm giving you the cartition coordinate system that will be much more easier to understand what exactly we

need to do. So to understand this I'm drawing the cartitionian coordinate system and I'm This point what is this point called in the cartition coordinate system if you are considering this as positive x this is negativex positive y and negative y what you consider this point origin okay so I'm considering it as origin but what exactly I'm going to say I'm I'm saying that consider origin as pole for your mirror like if you have concave mirror the pole should coincide with origin. If you have a convex mirror again the pole should coincide with your origin.

Now You need to measure the distance. For example, I'm having a object on the left side because the incident ray will be going from left to right. So this distance will be on the negative x-axis but the height of the object will be considered as positive because it is on the positive y-axis. If the image is formed on the right side, so I need to if I need to take the distance that will be on the right side on the positive x-axis from the origin or pole. So it Will be considered as positive distance. But

when the image is downward which is going on the negative y-axis it will be considered as negative. So these are the five points which what we need to you know understand. First of all the object will always be on the negative. So object distance will be negative for real inverted image. So real real images will be forming on the left side. So again the real inverted images will be considered as negative. For virtual erect images which which is generally going to be on the right side. So that will be considered on as positive. The height

of the object will be considered as positive because it is already on the exactly parall to the y positive y-axis. And real inverted images the height of the image will be considered as negative because it is parall to the negative yaxis. That was the significance of sign convention and by using this sign Convention I can guarantee you the numericals are never going to be incorrect. Use this sign convention your numerical will be absolutely you will be giving you if you are providing the you know correct values place the value along with the proper sign you

will be getting the correct answer for sure. Moving ahead to the mirror formula. From here the numericals are going to be appear. So just be ready for that. But before That we need to learn the formula. The mirror formula says if a it is the relation between object distance, image distance and focal length where U is the object distance. Object distance will be represented by U. Image distance will be represented by V and small F. We already know that this is nothing but focal length. So by using this the relation between all three will be

written as 1 by V + 1 by U is equals To 1 by F. So that is important magnification. So that is applicable first of all for the concave and convex both. So you don't need to worry you do not need to remember the separate formulas for the concave mirror and convex mirror. Talking about the magnification see generally when I'm talking about the magnification people says it is just going to magnify the image. It is exactly not like that magnification is the reference or it Gives the ratio or a kind of clarity to us whether

the size of the image is been increased or decreased. So magnification so that is the ratio of the height of image to the height of object and also it is image distance by object distance with a negative sign. Image distance by object distance with a negative sign. So that is called the linear magnification. So that is the formula for the linear magnification that is m= to minus v by u equals to height of the image to the Height of object. That's how you need to write it. There are few notes which I mentioned over here

like if magnification value is one what does that mean? It means that the size of image and the size of object is equal of course and the image distance and the object distance are also equal. What if we have a virtual image? In that case if magnification given as positive that means for sure the image is virtual. But If the magnification given as negative be it any value the image will be real and inverted. We just need to clear that in our mind. What does that mean? Like m is positive that is virtual and direct.

If m is negative, it will be real and inverted. Is that clear? Now we will be starting refraction. So what is the refraction? So just make the another segment of the refraction of light. And the refraction of light as per the definition it says the phenomena of Change in the path of the light as at it as it travels from one transparent medium to the other medium. So it it is very very easy to understand if the ray of light it is incidenting from one medium to another medium it deviates its path or it changes

its own direction. This is what the refraction is. So if you see if I have a glass slab and the ray of light is coming from the air medium it will be bending towards the normal That is called refraction. Similarly, the ray of light which was traveling into the glass medium and entering into the air medium, it will be bending away from normal. In class 8, you must have you know gone through with this refraction cases. The thing is why all this refraction phenomena occur. The only reason is the refraction occur because there is a

change in speed of light in the different medium. In air medium generally the speed of light is Quite higher nearly to the speed of light in vacuum that is 3 into 10 ^ 8 m/s. Whereas in the other medium be it water or glass the speed of light goes decreasing because the particles are quite closer. As a result the light ray becomes deviated maybe towards the normal or away from normal and this phenomena is called refraction of light. Let's talk about few other thing what are the causes. So if we talk about the basic concept

so the light is traveling From here to here you must have observed there if you take a glass filled with water and if you're inserting the pencil into it you you must have seen there is a slight bend in the pencil it is because of the refraction of light. That is the only reason how it happens which I'm going to tell you like in few minutes you need to wait and moving ahead what are the laws we need to understand the laws of refraction. So [snorts] in the reflection there were Two laws. Similarly in the

refraction of light we again have two different laws. The very first law which is equal to the second law what we have studied in reflection of light that is the incident ray refracted ray I'm repeating once again refracted ray and normal all lie on the same plane at the point of incidence whereas second law says the ratio of sign of angle of incidence to the sign of angle of refraction is a constant. So this is giving us some trigonometrical term which clearly says that the sine of angle I by sin of angle R that is

nothing but constant. So if we divide these values sin I upon sin R we will get a constant. Let me give you a trick. This this is going to give us the refractive index. Refractive index. What is the refractive index? Many students are getting confused like sir I'm I'm having so many doubts in Refractive index. So one by one we are going to discuss. But yes there is one more formula for the refractive index that is sin I upon sin r. But according to the snail's law sin I by sin r is constant and this

is what we are going to use in the coming slides. So if we see the refraction through a glass slab what is going to be happen? So let's consider we have a glass slab PQRS and a ray of light AB is traveling from air medium and entering into the denser Medium. So clearly I'm saying the light is traveling from rarer medium entering into the denser medium again entering into the rarer medium. This is how the light is going to travel. Correct? So when the light is traveling from air medium when it enters into the denser

medium it bend towards the normal and when it again enter into the air medium it is moving away from normal because the ray of light again coming back into the same medium from Air to air the light rays will become parallel how I got to know let's have a look if I extend the original incident ray Just like here I did if I'm going to extend the original incident ray here. So I can see these two lines are parallel to each other and these two parallel lines gives us some value that is called lateral displacement.

What it is going to give us lateral shift or lateral displacement. lateral displacement Or lateral shift. According to the definition of lateral displacement or lateral shift, it says the perpendicular distance. Which distance? The perpendicular distance between emergent ray. emergent ray and original incident ray. Original incident ray. So it if we take the perpendicular distance if we measure it you will get The perfect value of the lateral shift and this is how we calculate the lateral shift value. I hope you got this. Let's talk about the few other terms which are required that is one incident

ray you are familiar with refracted ray you know that but once the light is incidenting refracting and again coming back into the same medium the light ray becomes emergent ray and the angle between them becomes angle of incidence angle of refraction and angle of emergence. Amazingly this angle and this angle will be equal to each other. We need to apply simple mathematics in that if we see this is forming alternate interior angle. So it won't it is you know already been mentioned as R and Rdash but they will be equal to each other if it

is a perfect rectangular glass slab. So this is how the refraction in a glass slab occurs. Moving ahead. So if you look into the absolute refractive index. So what does the Absolute refractive index mean? Absolute refractive index define the ratio of speed of light as compared to the vacuum or air. So generally we we consider it as vacuum because the speed of light is maximum in vacuum. So refractive index will be defined as the ratio of speed of light from vacuum to the medium. So let's consider the light is traveling from vacuum entered into the

medium. So what will be the refractive index in that case? So that will be the speed of Light in vacuum to the speed of light in medium. That can also be write it as like refractive index is equals to C by V. This is another way of writing the value of refractive index in short. That is absolute refractive index. What is the other kind of refractive index? that is called relative refractive index. So when a light is traveling from one medium to another instead of vacuum or air. So in that case we call it as

a relative refractive index because there Are two different mediums having different densities and different refractive indexes. So in that case we call it as relative refractive index. So how we calculate the value? So the writing method that is refractive index of two with respect to one that can be writed as refractive index of one by refractive index of two that is you know uh the method number one of writing or to get the value of refractive index of two with respect to one you need to Write n_sub_1 by n_sub_2 but we studied that is n_sub_1

means the refractive index of medium one with respect to vacuum. So we know the formula as per the absolute refractive index that can be writed as C by V_sub_1 and N_sub_2 can be writed as C by V_sub_2. If I'm going to cancel out the value of C, I will be getting V_sub_2 by V_sub_1. So I have got the refractive index value as N_sub_1 by N_sub_2 or it can also be writed as V_sub_2 by V_sub_1. So that is the formula for the relative refractive index. Apart from that as per the notes. So you can write down

a note point. So in the note point what we have got the refractive index can be writed as C by V. Refractive index of 2 with respect to 1 can be writed as N_sub_1 by N_sub_2 that is equal to V_sub_2 by V_sub_1. Refractive index can also be represented as sin I upon sin R. And one more thing in case if the question being asked when there Is a depth of water being given and they're asking about the apparent depth or real or apparent depth of being given you need to calculate the value of refractive index.

In that case the value of refractive index will be real depth by apparent depth. I will be giving you an example to make you understand how we can calculate the value of refractive index or the given real depth or apparent depth according to this. So this is the note point you Need to consider this in mind while solving numericals that will fulfill your purpose completely. This is going to fulfill your purpose. Don't worry about that. So now some applications. So when you see the pencil what exactly is going to happen? Let's have a look. So

the pencil is completely straight like that. That is the pencil. But when we are looking from our eye, we see a straight line. So the light which is traveling From the bottom will be bending away from normal like that. This is going to bend. But because we see straight away it seems that the bottom surface is going to be a little bit higher at this point. This is how the pencil seems a little bit you know tilted or slightly bent. That is the only reason one of the reason why the uh the base of the

swimming pool appears to be higher. That is the reason correct. So water tank appears shallow and less deep than it's Actually are. The only reason is the light which is traveling from the bottom point of the water tank. The light which is reaching to us we are actually looking into the straight path. So it seems that the base image is going to be appear at this height. So at this height so we feel that the base is you know quite higher when you are going for the swimming you know you you feel that okay okay

I can I'm able to swim in in such depth but once you are going to jump Into it you realize the water is too deep or the swimming pool is too you know at a at a great height or depth and where I'm going to be you know drown inside it so you need to be very careful when you look into any river when you look into any swimming pool or water tank tank before jumping. If you want to analyze the correct depth of it, you need to watch, you need to see it very normally

because when the light is traveling straight away, no refraction Will occur and you would be able to see a actual depth of the water tank or swimming pool. So once you are looking from some angle, you won't be able to judge the actual depth of the water tank. What else? So now we are having few numericals based on the refractive index. So we are going to do that very interesting part. Students are confused in that. So don't worry we will be solving every single numerical with a proper you know the Channel or the proper guidance

of the board examination. This is what I'm going to give you again. So light travels through water with a speed of 2.25 into 10 ^ 8 m/s. So the speed of speed of light given that is in water given as 2.25 into 10 ^ 8 m/s and the speed of light in vacuum already given as that is C is equals to 3 * 10 ^ 8 m/s and I need to calculate the refractive index of water. This is what I need to calculate. So we know the Formula to get the answer. So refractive index of

water with respect to vacuum that is C by V. Yes. C by V. You need to write the answer in the comment if you get the answer before me. So refractive index of water that is 3 * 10 ^ 8 over 2.25 2.25 25 * 10 ^ 8 10 ^ 8 and this is going to cancel out and on on solving you will be getting the value of refractive index as 4x3 or if you want to keep it in decimal then it will be 3 By 2.25 what will be the unit of refractive index? So

there is no unit for the refractive index because m/s and m/s will be going to cancel out each other. So there will be no unit for the refractive index. It is just a fraction by which the light is going to bend. So that was the you know the kind of numerical asked on the basis of refractive index. What is the other kind? So the other kind is the light travels from rarer medium one. So they Clearly mentioned we have a medium one which is rarer, medium two which is denser. Of course the angle of incidence

and refraction are respectively given. So in the given data I'm going to write angle I given as 45° angle R given as 30°. We need to calculate the refractive index of second medium with respect to the first medium. Okay. So you know the formula in notes I already gave you. So the refractive index of two with respect to One that is equals to sin I by sin R. Let's put the value. So this is going to be sin 45°. This is going to be sin 30°. The value of sin 45 is correct 1x 2 1

by under 2. And the value of sin 30 is 1x 2 or 3x2 correct 1x2. So this is going to be 2 by 2. This can also be writed as square roo of 2 square roo of 2 by square roo of 2. And finally the value of refractive index which is which I am going to get That is under root2 and 1.414 can also be written the value of square root of two. So very easy to solve the numerical based on the refractive index. You only need to understand what exactly they are going to ask.

What of what the refractive index with respect to what once you got to know you will be able to crack every single numerical. What is the other type? So now the pond of depth 20 cm is filled With water. So now they have given us the depth of the water tank. So we are given with real depth which is nothing but 20 cm and the refractive index already given to us that is given as 4x3 appear they are asking for the apparent depth of the tank apparent depth you can even consider it as no problem.

So now for solving the refractive index is equals to real depth by apparent depth. This is what I told you right. Real depth by apparent depth. Let's put the value. So you will be getting 4x3. Real depth is 20. Apparent depth will remain as it is. So apparent depth will be equal to let's do the cross multiplication that is 20x 3x 4. On cancelelling I will be getting five and the value of apparent depth is nothing but 15 cm easy. So these are the kind of numerical generally in the refractive index they are going to

ask. Now what let's talk About the spherical lens. What are the spherical lenses? What kind of spherical lens generally the board examination will be going to ask? So according to our syllabus we are having two different categories of lenses. But how actually they are going to be you know made or how they look like. So we need to understand this. So in the spherical mirror we have considered only one spherical surface. Whereas in this we are going to take two spherical Surfaces. So if you see there are two surfaces and in this if I'm going

to cut this much of portion I will be getting one lens. Similarly if I'm going to place two different disputes in such a manner this surface will be considered as the lens which is of same material. Amazingly for this particular surface, please be concentrated at this part. For this particular surface, this will be the center of curvature. And amazingly, for The other part, this particular surface, C2 will be the center of curvature. So, can I say that there will be two center of curvatures will be given for any spherical lens. The answer is yes. There

will be two center of curvature. But exactly what we are going to consider. So we will be considering one primary center of curvature and another will be secondary on the basis of sign convention and rules. We will be defining one of them. So just please be Patient wait for a while you will be getting all the answers. Similarly for this surface this will be the center of curvature and for this surface this particular one will be the center of curvature. So in both the cases you will be getting different center of curvatures. So let's have

a look. So this is this [clears throat] kind of lens. If you are getting this kind of lens this structure that means the name of the lens The name of the lens is nothing but convex lens whereas this lens is concave lens. How you would be able to identify generally in the biba or practicals you know the examiner will be going to ask how would you identify it is a convex lens or concave lens if you are blindfolded. So the answer is very simple if you touch and feel it it feels Like it is thick

in the middle and thin at the edges. That is also the definition of the convex lens. It is thick in the middle and thin at the edges. Whereas concave lens is thin from the middle and thin at the edges. This is how you would be defining convex and concave lenses. We need to understand different terminologies based on the lens. So basic terms will be first we have the aperture. So if I have the diagram I have already drawn it. So if I Have a diagram so this particular dotted line this diameter will be considered as

the aperture. So aperture is nothing but the diameter of the circular edges of the lens. So this particular line will be considered as aperture. very last less asked question what is the center of curvature. So center of these spheres are called center of curvatures. I have clearly mentioned earlier also in the previous slide that there will be two center of curvature But we need to decide which one will be the primary one. Talking about the principal axis. So just imagine you are having one thing. Okay, one one one thing this particular center will be considered

as the optical center represented by O. In the spherical mirrors we were having pole. In a spherical lenses we are having optical center which is in the middle exactly in the middle somewhere in between the lens that will be the optical center. And Optical center is something from where if the light ray is going to incident it won't be deviated. It will pass undeviated. The definition says that. So the principal axis if I'm going to make a imaginary line passing through both the center of curvatures and optical center that will be your principal axis. So

imaginary line passing through the center of curvature of the two surfaces is called principal axis of the lens and optical center is Is a point on the principal axis of the lens such that the ray of light passing through it goes undeviated. That is the definition. So and one of the rule that will that is going to help us what is the principal focus for the lenses. So in case of this in case of convex lens if a ray of light coming parall to the principal axis it converges and meet at a point that is

called focus. It converges and meet at a point that is Called focus. And because this is going to converging this lens is known as converging lens. This lens will be known as converging lens. Whereas if a ray of light is incidenting or appear to incident on focus after refraction it will be going parall to the principal axis. This is how you would be able to find the point principle focus for both the lenses. Correct? Now, rules. What are the rules? Very very Important. So, rule number one. If a ray of light coming to the principal

axis after incidenting on convex lens, it will be passing through focus. Yes, we know that the light ray always passes through the focus when it is coming parall to the principal axis. But the only thing is it it need to be converge on the other side because it is not reflection it is refraction. So it will be going on the other side. This is how we need to Define it which clearly says that according to this the focus will be on the right side of the convex lens and because the focus is on the right

side the primary center of curvature will also be on the right side of the lens. Got it? Point number two. If a ray of light passing through the optical center, it will be going undeviated. So the straight line will be passing through it. again the refraction occurs and the focus will be on the right side. Rule number three, if a ray of light coming to the object and passing through the principal focus. So consider the ray of light passing through the first focus of the convex lens after refraction it will be going parall to the

principal axis. It is exactly like the rule number one that we have studied two slides back. Correct? So this is how it is going to pass. These are the major rules which are required to form the images and we will be getting the complete Image formations of the lenses. So let's begin. So for the concave lens what are the rules? Okay I need to mention the rules for the formation of images. So if a ray of light going parall to the principal axis it will be deviated. If it is going to deviated that will be

our diverging lens as I mentioned earlier also. But it will appear to meet at a point that is called focus. So now you know you need to put more concentration. The focus will become on The left side. If the focus is on the left side for the conve concave lens the center of curvature will also be considered on the left side of the concave lens. So for the convex lens the focal length will be on the right side. Center center of curvature will be on the right side primary one. And for the concave lens focal

length of focus will be on the left side as well as the center of curvature will also be on the left side of the lens. Next rule. So if A ray of light if a ray of light passing through the optical center it will be going undeviated as shown like the previous one. So now the formation of images for the convex lens one by one we are going to see for the convex lens we are having six different cases same as the image formation we did in the concave mirror the rules will be same the

formation of images will be same only the thing is the refraction is going to occur and there will not be any Reflection there will be a complete reflection which you are going to see so let's have a look so the if the rays of light coming parallel like that. So after refraction it will form an image. It will be forming an image at focus. So now you can see here we have the a-b dash. So I'm going to write for the image if the object is at infinity the image will be at focus on the

right side. Second, it will be highly diminished or point- sized. And because the light rays are actually meeting, the nature will be real and inverted. I hope this is very loud and clear to you. I hope this is very loud and clear to you. Right? Rule number two. If if when the object is beyond center of curvature. So I have kept the object just beyond 2F. Object will always be kept on the left side. So I have kept the object AB just beyond Center of curvature on the left side. So the rays of light passing

after refraction after refraction it will be passing through focus. So it is going like that. And the light ray which was passing through the optical center will be going undeviated. And we have observed this is and we have observed we have got an image at this point a dash bdash. I'm naming it as a dash bdash. And I can Clearly see that the image is popped between focal length and center of curvature. So I'm going to write about the image when the object is beyond center of curvature image will be between focus. So if I'm

going to write about the images. So image point number one. So when the object is beyond center of curvature image will be between center of curvature and focus. Second it will be diminished or smaller than the Object. And talking about the nature it will be real and inverted. So that was the rule number two for the formation of image. moving ahead. If I'm going to, you know, moving slightly, you know, moving just at the center of curvature, what is going to be happen? So, I've kept my object at center of curvature. I've just I have

just kept the object over here. What I'm going to get? Let's follow the rule. If the ray Of light going parall to the principal axis, it will be passing through focus. Perfect. And if a ray of light passing through the optical center, it will be going undeviated. Again perfect. Now I have got an image that is exactly at center of curvature but on the other side I have got the image but on the other side where at center of curvature so let's talk about the image. So image will be at center of curvature on other

side Not on the same side or right side you can say. Second, the size of the image will be same size as object exactly of the same size. And point number three, the nature of the image will be real and inverted. Once again, that was the rule number three. Now, if I need to move again slightly my object, now the place of the object will be between F and 2F on the left side. So I have kept my object at this point. Now I'm going to follow the rules. And What are these? Parall to the

principal axis passing through focus. And if a ray of light going through the optical center, it will be going undeviated. And finally the image will be formed where? At this point and this point seems beyond 2F on the other side. So let's talk about the image. So when the object is capped between F and 2F on the left side image will be forming beyond Center of curvature. The size will be enlarged. Now it is going to enlarge itself and again the nature will be real and inverted. The reason is very simple because the light rays

are actually meeting. Case number five. Now I'm going to keep my object at focus. So I have kept the object at focus. So this is my object and following the rules. So I can see that if the rays of light going parall to the principal axis, it will be passing Through focus. And the light ray which was passing through the optical center, it will be going undeviated. Amazingly these two light rays, the two refracted rays are not going to incident. They are parallel to each other. And we know this concept from the spherical mirror story.

If the light rays are parall to each other, they never meet. They will meet at correct at infinity. What will be the size of the image? That Will be highly enlarged. Highly enlarged. What will be the nature of the image? Only a screen can obtain the image of it. So it will be real and inverted. And the last case when the object will be placed between optical center and focus and you know what is going to be happen. So if I'm going to place the object here I'm going to follow the rules. So in that

case if the ray of light going Parall to the principal axis it is passing through focus like that and through the optical center it will be going like that. Now these two light rays are not parall to each other. These two light rays are not parall to each other. They are actually away from each other like that. They are not parallel. They are slightly away from each other. So I need to extend it back to get the value or to get the complete image. So I'm going to extend it and wherever they Are going to

slide the image is going to be formed. So I can see that here I have obtained my image a-bash which is on the same side that is on the left side because the images image is formed due to appear to meet category. So that will be virtual and direct. So I'm going to write image form on same side as object. Same side as object. Point number two, size will be enlarged. Point number three, the nature will be Virtual and erect. virtual and erect. This is what exactly happened almost you know the image distance or the

position of image the nature of image about the size this is exactly we studied where in concave mirror which is going to be happen in convex lens. So it will be quite easy if you know the concept of spherical mirrors it will be quite easy to understand the concept of spherical lenses but while reading numericals you need to be very careful whether the Question is given for the convex mirror or lens. So just be concentrated on that. Now talking about the concave lens. So we have taken if the object is taken at infinity. Let's consider

the object is at infinity. So the light rays which was emerging from that particular object will be parall to the principal axis. So it will be going like that and after refraction it will be diverging because it is a diverging lens. So now you can see the light rays have been Diverted. Now if the light rays have been diverted they are never going to meet in that case I need to take it back and the image will appear at one point and this is how we can calculate or evaluate the value of focus in the

concave lens we know that so I'm going to write image will form if the object is kept at infinity image will be formed at focus size will be sized or highly diminished. You can write third the nature will be Virtual and erect. The last case for the formation of image of the spherical lens in fact that is the second formation for the concave lens that when the object is kept anywhere anywhere between infinity and optical center. So I have kept my object at this point that is AB. Now the light ray going parall to the

principal axis will be diverted like that. And the light ray which is passing through the optical center will be going Undeviated. On emerging back, on emerging back the light ray will be incidenting at this particular point. At this point here, the image will form. At this point, the image will be forming. So I can write it as image will form between optical center and focus. The size will be diminished. And nature will be virtual and erect. Virtual and erect. That was too easy, right? Because you Were already familiar with the spherical method. Let's conclude it.

So for the formation of image, we are talking about convex lens. [cough] So when the object is at infinity image will be at focus highly diminished real inverted beyond 2F image will be f and 2f between f and 2f diminished real inverted at center of curvature image will be at center of curvature same size real inverted between focal length and Center of curvature image will be beyond 2f magnified real inverted at focus infinity highly enlarged or highly magnified real inverted and between O and F on the same side magnified virtual and direct. Same as for

the concave lens. Object at infinity image will be at focus highly diminished virtual and erect and anywhere between optical center and infinity image will be between optical center and focus Diminished virtual and erect. So that was the end of the formation of images by the spherical lenses. Now we will talking about spherical lens and its formula and sign convention. Sign convention will be exactly same that you studied in the spherical mirrors. The only difference will be the all the distances will not be measured from pole. The distances will be measured from optical center because in

lens we are having optical center instead of Pole. So point number one it says the all the distances will be measured from the optical center. Incident ray will be taken from left to right. Distances towards the right will be considered as positive. Towards the left considered as negative. Height above the principal axis will be considered as positive and height below the principal axis will be considered as negative. No changes required. Only you need to replace pole with the optical center and The sign convention for the spherical lens is done. Now talking about the lens formula,

we studied about the lens formula, spherical mirror formula which was giving the relation between U, V and F that is object distance, image distance and focal length. Similarly, the lens formula also give us the relation between U, V and F. But there is only one change required. Instead of using positive sign, I will be putting a negative sign in between. And the Formula becomes 1x v - 1x u is equals to 1 by f. And the magnification formula that is going to be n equals to height of image by height of object that is equal

to image distance by object distance without a negative sign. In the mirror it was a negative sign placed where after you know before v and u it was already given it it was minus v by u here we are not required to put a negative sign. So that is the formula for the magnification. What Else? Let's talk about the last con concept of this chapter that is the power of the lens. So power of the lens is defined by the by the ability of the lens to converge it like how much it can converse the

light ray or diverse the light ray. This is what the power of the lens is or in a simple language or in a oneliner or objective type this is how it appears. The power of lens is nothing but the reciprocal of the focal length and this is how it seems like the focal Length is equal to one by focal length. But you need to keep one thing in mind it has to be in meters. But if the focal length is given in centm just change your formula in you know a bit you can write it

as 100 by focal length in cm or otherwise you will have to convert the focal length from cm to meter then you can use this formula. I hope you got this concept. Now what if I'm going to combine you know few uh lenses all together what is going To be happen what kind of power is going it is going to be so it's it's a very simple thing I need to evaluate the individual powers of that those lenses once I get it do the algebraic sum you just need to add them with their signs whatever

you are getting a positive or negative depending on the value of power and focal length on the basis of that you need to put the value of power that is P equal to P1 + P2 plus P3 and so on whatever number of lenses you are going To use. Let's do one thing. Let's solve some numericals and get the value to understand this. So first of all we are provided with a 2 cm tall object is placed perpendicular to the principal axis of convex lens. Now one by one we are going to solve this. We

are provided with a convex lens. The object distance given object height given to us that is 2 cm of a focal length 10 cm. So the focal length given as 10 cm. Remember one thing where the Focus focus of convex lens occurred. Was it on the right side or left side? So imagine this is our f convex lens. So when the ray of light coming parallel when the ray of light coming parall to the principal axis it converges and this is the point where I will be obtaining my focal length which is on the right

side of the lens. So I will be take considering it as positive. Now [clears throat] we are given with the distance of the object from the lens is 15 cm. So the object distance that is U given as 15 cm and object distance will always always always and always will be negative and will be placed on the left side of the lens. Now I need to calculate the value of position of the image, nature of the image, size of the image. This is what I need to calculate. Also they asked me to find out the

value of magnification. Now I understood what exactly given to us what I need to find. Let's put the Value and do it. First of all I'm going to apply lens formula that is 1 upon v minus 1 upon u that is 1 upon focal length. 1 upon v I'm not familiar with negative sign object distance given that is -15. So I'm going to put the value focal length given as positive 10. So I'm going to put positive 10 - will become + and this + 1 by 15 will go on the other side and positive

1x 15 is going to be - 1 by 15. So 1 upon v by taking lcm I will be Getting 30. So this is 3 - 2 and the value of 1 by v is 1 by 30. Let's do the reciprocal and I will be getting the value of V as 30 cm with a positive sign which means the image will be forming on the right side. That is the only reason the image distance will be positive. Now moving ahead of object given I will be using the formula for the or the relation my bad

the relation between hi by ho that Is equals to v by u so height of image I need to calculate height of image given as two value of image distance given as 30 object distance is -15 now solve this 15 and 30 will be get getting cancel out. So, hi is equals to 2 will be going on the other side. This is going to be -4 cm. Part number C that is magnification. To solve this, you can either go with hi upon ho or you can even do v upon u. In both ways, you would

be getting your Answer. So magnification height of image is nothing but minus4 height of object given as 2 and the magnification that I have got that is -2 and because the magnification value is negative which means the image is real and inverted. If the value of magnification is positive that will be virtual indirect because since we have got the negative value it will be real and inverted. Moving ahead to the next question it Says now we have a concave lens which has a focal length of 15 cm and focal length for the concave lens will

be considered as negative. So I'm writing as -15. I'm writing -15 value. At what distance should an object be object from the lens be placed so that it forms an image 10 cm from the lens. So it says the value of V is -10 cm and the value of U I need to calculate also find the magnification. This is what The image in the concave lens you know generally occur on the left side. So I have kept with a negative sign. So I'm you know uh in the given data I have put all the correct

values. I only need to solve this. So let's do it. Part a 1x v - 1 by u that is equals to 1x f. The value of v given is -10 minus 1x u and that is -15. So I'm going to take - 1x u on the other side and - 1x 15 on this side. So what will be I getting 1 upon u that is equals to - 1x 10 + 1 by 15. Let's take I will be getting 30. This is -3 + 2. So 1 by u that is -1 by 30. And

the value of u is nothing but -30 cm. [clears throat] So to get the value or the to get the image at 10 cm if the concave lens is of 15 cm the object should be placed at 30 cm. Talking about the magnification. So magnification formula is nothing but v upon u. So the v value given as -10. The value of u I have just obtained as Minus30. So the magnification is 1 by3. Done. In concave lens in both the cases in both the cases whatever formation of image we have seen if the object is

at infinity image will be at focus point. S that means the magnification will be less than one. Even in the second case, the image size was diminished or smaller than the object which means the magnification value will again be less than one and This is what I have got 1 by 3 which is less than one and it clearly states that the image size will be smaller. Moving ahead to the other question now again we are provided with a f concave lens of focal length 25 cm convex lens of focal length 20 cm. Now we

are provided with two different lenses. So concave lens so I'm writing it as f_sub_1 considering lens one whose focal length given as 25 cm as it since this is a concave lens it will be considered with a negative sign For a convex lens it will be considered with a positive sign and the value given as 20 cm placed in contact with each other. So both the lenses are capped along each other. So this is this is concave lens and this is convex lens. They capped along with each other side by side. What is the power

of this combination? So they are asking about the power of the combination also calculate the focal Length of the combination. Quite easy we know the formula for the to calculate power. So I'm going to calculate individual powers. So power one for the concave lens this is going to be 100 by focal length 1 that is 100 by -25 which is nothing but -4 diopter the unit for power is nothing but diopter represented by capital D similarly power two that is equal to 100 by focal length 2 this is going to be 100 divided by 20

which is nothing But Five diopter the combi the combination power is going to be P1 + P2 let's put the value so -4 + 5 and power for the combination will be nothing but one diopter what I just did calculate individual powers add them get the value now I need to calculate the focal length for the combination so I will be using the power for the combination only. So in part B power of combination that is equal to 1 by focal length of the combination. So Focal length of combination that is 1 by 1. So

this is going to be 1 m because I'm considering one by focal length that means by default it has to be in meters. If I'm going to consider 100 by focal length, the answer will be 100 cm. So in both ways you can solve and get the value. So that's your answer. I hope you got the concept of you know the combination of cell how to get the Power and focal length. Again we are provided with the combination numerical. So a convex lens of focal length 20 cm is given. So let let's write the given

value. to the focal length of the con convex lens that is 20 cm and in contact with concave lens whose focal length is 10 cm with a negative sign. Find the focal length and power of the combination. So power of combination focal length of combination to solve this. So in first case power for one that is 100 by focal length 1. So that is 100 by 20 which is going to be 5 diopter. Similarly for power two that is 100 by f_sub_2 which is going to be 100 divided by minus 10 which will be -10

diopter and power for the combination how much it will be 5 + 10 or 5 - 10 correct it will be 5 - 10 so power for the combination is nothing but 5 diopter absolutely right answer guys absolutely right answer but I also need to Calculate the value of focal length for the combination. So power of combination that is equal to 100 by focal length. Let's let's try for the uh to get the value in cm in cm. So focal length for the combination that is equals to 100 by -5. So which is going to

be this is 20. So this will be -20 cm. Now this is your answer. That's how you solve numericals based on the power of combination of lengths. Now Important segment that is P YQ's what kind of question been asked in the previous years in the board examination. Let's have a look and let's try let let me see just write down in the comment box how many question you made right how many questions you have attempted or not attempted separately so that we will be analyzing our preparation on the basis of whatever concept we have studied

so far. So moving ahead to the very first question. It says an object is placed at A distance of 30 cm from the reflecting surface of a concave mirror of radius of curvature 40 cm. The image formed is to get this I need to write the given values. The object is placed at 30 cm. The radius of curvature for the concave mirror given as -40 cm. Which means the focal length will be -30 cm. If the focal length is 30, center of curvature is 40. Object placed between C and F. Object placed between C and

F. Now remember this case object is between P And F. Now remember the image. Image will be beyond C size. What about size? Enlarged, nature, real and inverted. Now look at the options. We are provided with virtual and magnified, virtual and diminished, virtual and magnified. What re sorry real and magnified real and diminished. So we are getting enlarged value that Means magnified. So option number C will be the right answer. So this is the reason you know the formation of images and the rules for the formation of images are highly required necessary very very important

for you because such kind of question already been asked in the previous years. Moving ahead for the next question which says a students want to obtain an erect image of an object using a concave mirror of 10 cm focal length. So it is a Concave mirror. So this is my concave mirror. I'm just drawing the principal axis. This is pole. This is pocus. And to obtain the erect image, I need an erect image. And erect image can only be obtained when the object is capped between pole and focus. So I need to place the object

over here to get a image on the either side which whose nature will be virtual and direct. So what exactly they are asking? What will Be the distance of the object from the mirror less than 10 cm or focal length? 10 cm between 10 and 20 more than 20 I don't think this will be taking more than 10 seconds to make it right so the object should be placed less than 10 cm which is less than its focal length and we will be getting the virtual end erect value next question says the image of an

object placed in front of a concave mirror of focal length 15 cm is of the same size The distance between object and images same size. The only case the only and only case where the image will be forming at the same place or of the same size where the object distance and image distance will be equal to each other that is center of curvature. So the object placed at center of curvature image obtained at center of curvature. Same size same distance. Distance between object and image. So if this is [clears throat] your object, This is

your image. No distance required. A B C D. Answer is D. Zero. Correct answer. So this question been asked in 2022. Moving ahead to the next question. At what distance from a convex lens should an object be placed to get an image of the same size as that uh that of the object on the screen really they again ask the same question like just after a year so where I need to put the image at center of curvature so let's have a look beyond twice the focal Length that means beyond center of curvature at principal

focus No no twice the focal length at center of curvature the eye should you know give a spark to see this option let's have a look to the other one between optical center and principal focus not at all so the option number C is the right answer 2023 they asked like when an object is placed beyond 2F of a convex lens the nature of the image formed is really beyond center of curvature we know that We earlier discussed about it beyond 2F beyond 2F let's have a look so if we go beyond 2F image will

be between F and 2F it will be diminished real and inverted amazingly we would be able to solve this question so the image will be real inverted diminished done real inverted diminished virtual cannot be made. Real, inverted, magnified, magnified cannot be you know considered because once you are moving Towards the optical center after center of curvature then the image gets magnified. So magnified will not be the solution. So all three options are correct. The first option is absolutely right. Moving ahead to the another question. Last year this question been asked in the previous year. An

object of height 4 cm, okay, is placed at a distance 30 cm from the optical center of convex lens of focal length 20 cm. Now they ask some Serious thing, okay, let's do it. What are the given values given to us? The object of height that means the height of the object given to us that is 4 cm placed at a distance 30 cm that means the object distance given as -30 cm from the optical center of convex lens of focal length 20 because it is a convex lens the focal length will be considered as

positive. using lens formula the distance between image from the optical center they asked me to find the image Distance okay do it so 1 by v minus 1x u that is equals to 1x focal length so 1 by v I need to calculate object distance already given that is -30 and the focal length given as 20 so minus minus is plus if I'm going to take it on the other side this is going to be negative so 1x 20 - 1 by 30 let's take the LCM I will be getting 60 so this is 3

- 2 so 1 by v will be equals to 1 by 60 and the value of v is coming out as + 60 cm That is the value of image distance from the optical center second height of the image formed Okay. So we know the relation hi upon ho that is equals to v by u. So the height of image is unknown. Height of object given as four. Image distance 60. Object distance minus 30. Let's cancel out. You will be getting two. And the height of image that we have got that is -8 cm which

clearly states the image will also Be real inverted because the height came out as negative. That means it is below the principal axis parall to the negative y-axis. Done. Okay. So, okay. One last question. The value of magnification M for lens is -2. Using cartician sign convention and considering that object is placed at distance 20 cm from the optical center of this lens. The nature of the image formed is amazingly in the concave lens. Concave Lens do not have this capability to produce any real image because the magnification given is -2. So this cannot be

possible. We are provided with m = to -2 and and it gives the considering object is placed at 20 cm from the optical center nature of the image formed option number one. Okay, let's do one thing. I'm going to write it here. Since magnification is negative, the nature will be real and Inverted which can only and only possible if I'm going to use convex lens. Concave lens do not have this capability. The size of the image compared to the size of the object. This is what I need to calculate. So we are given with magnification

equals to minus2 that is equals to hi by h o. So height of image is equals to two times of the height of object. So the size of image compared to the size of object. So I can say that it is twice As of object. Height of image is twice of the size of the object. Talking about the position of image. position of image that means I need to calculate the value of V. For that I'm going to use h i by h o that is equals to v by u height of image and height

of object given as -2 that is equals to value of v which is unknown and the value of u given as -20 and the value of v is finally coming out on by doing the cross multiplication as 40 cm And the last sign of the height of image of Of course the sign of height of image is coming out as negative. No question required already. Already already you know it's it's already there. The height of object will always be positive and the height of uh image in that case will be coming out as negative. So

negative. So in this chapter what exactly we have covered? We discussed about the reflection where we covered the reflection through The reflection to the plane mirror. Spherical mirrors and the entire study about the spherical mirrors which includes types which includes formation of images. Formation of images which includes mirror formula. Then we switched to refraction refraction of light. Then we move to spherical lens Which include types again formation of image formation of images. What last? Lens formula and magnification. It it comes under this category lens formula and power of lens. These were the major topic that

we have covered. Apart from that we did so many numericals based on above category and at last we did PYQ's what kind of questions been asked in the Previous year. So that was the entire chapter or the oneshot of this wonderful chapter that is light reflection and refraction. You need to revise the concept. I will I would highly recommend not to revise too much of theory rather write down the theoretical question practice accordingly and apart from that solve as many numericals as you can and this is how you would be able to crack this chapter

and you will be getting 100% marks. I would also request you to Solve the DPP or the practice sheet. What what you are going to get our PW foundation English channel on the free batches and solve them, conclude them and achieve as many marks as possible. So, thank you so much. This is your teacher Vikas Agraal. Thank you so much guys. Have a great day ahead.