Light Reflection & Refraction Class 10: Complete Board Exam Crash Course

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 <unk>2

1 by under <unk>2. And the value of sin 30 is 1x 2 or <unk>3x2 correct 1x2. So this is

going to be 2 by <unk>2. This can also be writed as square roo<unk> of 2 square roo<unk> of 2 by square roo<unk> 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