With regard to ray optics, you should know what is reflection. Reflection means a ray of light is falling on a mirror. Reflecting surface is a mirror. Mirror will always be put like this. To show that one side is polished. So that your ray is coming like this. This is the incident ray. At this point of incidence, you should always have a normal. Normal perpendicular. And this is your angle of incidence. The incident ray makes with the normal. This is the reflected ray. It is getting reflected back in the same medium because it is a glass. This is angle of reflection which is again made with the normal. So this is how reflection happens. Reflections happen only the mirror surface. Mirror surface the reflection will happen. And you know this angle of I is equal to R. The angle of incidence should be equal to always angle of reflection. And the law Laws of reflection. These are 10 standard portions. Laws of reflection states that the incident ray, I will do it in red color, the incident ray, the reflected ray and the normal which are at the point of incidence. Normal is drawn at the point of incidence are all in the same plane. First law states that angle of incidence is equal to angle of reflection. The second law states that the incident ray, the reflected ray and the normal at the point of incidence all of them lie in the same plane. This is known as laws of reflection. In reflection if you see the ray is coming in air and going back into air. The same medium. So, you know speed of light in air is c. And it's also going back speed of light in air as C. The speed does not change. So let's talk about refraction. What is refraction? You need a medium, a glass slab or something. Okay, so then therefore when a light ray is entering from air, this incident ray, always remember at the point of incidence you have to do a normal. The light ray is entering from air into glass here. So it is entering from rarer medium into a denser medium which means obviously it will bend towards the normal. Always when it comes from rarer to denser, it bends towards the normal. Why? Because the speed will get decreased. Air is very fast. So, you know, here the velocity of light in air is c. But here, the velocity of the ray in medium is v. It's not c. It will be lesser. 3 into 10 to the power 8 is the maximum speed. meter per second is the maximum speed. Air le rukumadu apalaa fasta varudha, glass kular noraayum boodha, the speed will get reduced. And this phenomenon by which the light is travelling from one medium into another medium is called refraction. So what is the angle of incidence is the angle the incident ray makes with the normal. What is the angle of refraction? The refracted ray makes with the normal. So you understand. So in this, Here, you know the first law i is equal to r. The second law in the previous one is incident ray reflected ray and the normal at the point of incidence, all of them lie in the same plane. This is laws of refraction. What can you say about laws of refraction? You know refractive index is nothing but sin i by sin r. This law is called Snell's law. Again, these are all 10 standard basics. Okay, Snell's class states that the refractive index is the ratio of the sine of the angle of incidence to the sine of angle of refraction, correct? Sine i by sine r. You can also say, this is one way of saying, another way of saying refractive index is also speed of light in medium. Speed of light in vacuum by speed of light in medium. So speed of light in vacuum will be highest value. So denominator will always be lesser. So then therefore I will say refractive index of water is 1.33. Refractive index of glass is 1.5. Refractive index of air or in vacuum is just 1. Okay, something like that. Okay, so then there are refractive index for every medium, especially water and glass you need to buy harder. 1.33 and 1.5. Are you clear? So then... So, you know the speed of light in vacuum by speed of light in medium. Okay, so let me write this refractive index of water with respect to air. Water with respect to air. Light ray is coming from air into water. So, you are writing refractive index of water with respect to air and the value is 1 point. It is also mu of water. by mu of air. Mu of what is 1.33? Mu of air is 1. So, the answer is 1.33. Refractive index of glass will be with respect to air because it is entering from air to glass which will also be mu of glass by mu of air mu of glass is 1.5 mu of air is 1 so it is 1.5 so you will not detail that way you just have to know that this is mu of water mu of water knowledge with respect to air now in this okay so it is 1.33 and this is 1.5 this is speed of light in vacuum which will be v and which will be c vacuum means c you know by speed of light and medium which will be v So, you know this formula is c by v and you should also know that velocity is distance by time. 1 by t is frequency. Distance is lambda because it is like a wave, treated as a wave. You know that velocity is lambda into frequency. Do you know this? Distance will be wavelength and 1 by t is nothing but frequency. Reciprocal of time is frequency. So, therefore, you can write velocity as lambda into mu frequency, lambda into frequency. So, here it is lambda in vacuum into frequency in vacuum by velocity in medium, lambda in medium. frequency in medium. I am going to tell you something very important. The frequency does not change while it is travelling from one medium to another medium. Only the speed changes which means only the wavelength changes. speed changes now what changes only the wavelength changes the frequency does not change at all frequency does not change at all so then there is no frequency in vacuum and frequency in medium it's all the same so it's going to be the same the frequency in vacuum and frequency medium also is also vacuum same so it's going to be cancelled therefore you have another formula which is wavelength in vacuum by wavelength in medium is also refractive index therefore you understand that refractive index is inversely proportional to wavelength because wavelength and vacuum will be a constant value wavelength and medium is what is going to change so depending on this the refractive index and wavelength are inversely proportional are you clear so between red light Vibhji, you know Vibhji, red light and blue light. Let's see which has got a greatest wavelength. Red has got a greatest wavelength. That's why red is used as a signal. Because it can travel long distance and stop the vehicle. Whereas the green signal, blue is not green. But you know red has the longest wavelength. Blue has got the shorter wavelength. Red can travel longer distance because it has got a greater wavelength. Whereas blue has got a shorter wavelength. Do you understand? So, So wavelength of the red light is the maximum. So wavelength of the blue light is less. Okay it goes like this VibGyor where red has the maximum wavelength. It keeps on decreasing. Blue is here. Violet has the least wavelength. VibGyor returns like this where red has the longest wavelength. So then, therefore, you know that if you are talking about refractive index for a red light, Red light has got the greatest wavelength. So refractive index will be least for it. And the blue light is lesser wavelength. Therefore, it has got the maximum. So you should understand this line. Very important that the refractive index is inversely proportional to wavelength. Are you clear till here? Play it again and again. Become clear so that I don't have to do it now. So then this is what it is. These are the basics. Okay. So, I will just try the sum which is example 13 in Aurora. So, let us do example 13 in Aurora. A ray of light of frequency 5 into 10 to the power 14 hertz is passed through a liquid. So, you are passing this frequency into a liquid. The wavelength of the light measured inside the liquid is found to be inside the liquid which means wavelength in medium, wavelength in liquid is nothing but wavelength in medium is 450 into 10 to the power can you tell me minus, minus 9, minus 9 meter. What is given first? The ray of light with frequency. What is the meaning of frequency when passed through liquid? Frequency in the medium. Do you think there is something called frequency in the medium? I just told you frequency in air and frequency in the medium will be the same. Which means frequency in the medium is also frequency in air. General frequency. So, first clear it. Then they have given the wavelength in the medium. Is 415 to the power minus 9. Calculate the refractive index of a liquid. So, they are asking for refractive index. refractive index of the medium which is the liquid with respect to air And I told you the formula such as sin i by sin r. Snell's law we can't use it. There is no incident ray or refracted ray. I can't use that formula. What is the other formula I can use? Velocity of light in. vacuum by velocity of light in the medium. So you know that I can also write this velocity as wavelength in vacuum into frequency in vacuum by wavelength in medium into frequency in medium which is also frequency in vacuum. So the frequency does not change during refraction. Only the wavelength and the speed changes. Okay so then therefore what do you have you have wavelength and the wavelength in medium is there already given. and they have given the frequency in air frequency in air with with frequency medium is being given which is nothing but frequency in air with that you can find the wavelength in air because you want wavelength in vacuum which is nothing but wavelength in air you should remember this formula mu is equal to C by lambda You know velocity is distance by time. This mu is different from this mu. This is the refractive index mu. Don't confuse yourself. This is refractive index. This is frequency mu. Are you clear? The difference between this mu and that. This is refractive index. That is frequency. velocity is equal to distance by time which is nothing but v is equal to lambda into mu or you can also say c is equal to lambda into mu so if you want lambda you can write c by mu lambda is equal c is equal to lambda into mu so lambda is equal to c by mu so if you want to find the wavelength in air you can write wave the velocity in air by frequency in air so 3 into 10 to the power 8 by 5 into 10 to the power 4 protein you've calculated this answer you'll get that which is nothing but wavelength in vacuum or wavelength in air is nothing but wavelength in vacuum. So you will know this also you know this also so you can calculate the refractive index of the liquid. Are you clear? Another approach is C is above C The lower part is the frequency and vacuum So 5x10 to the power 14 Wavelength and medium is also given minus 9. This is 3 into 10 to the power 8. This is also right. This is also right. You can do whichever way. This is a brilliant way. In fact, because that saves me from lot of other workings. So, it is brilliant. So, try to continue doing like this. Think about your own options and do it. But this formula you should know. The formula is mu is equal to C by lambda. Mu is equal to C by lambda. So, now the question is only about find only the wavelength in vacuum. You cannot know this formula. suppose correct so other cancer one on the formula three of me partner correct on either 30 no so music will to see by lambda lambda is equal to see by me so with this you can find wavelength and any how you do is up to you i'm just teaching you one beef and nothing strikes you please do it are you clear with some very very important let's go to next question the light of wavelength 4500 Armstrong, 4500 Armstrong means 4500 x 20, see, minus 10 meter. In vacuum, so this is in vacuum. Enters a glass block of refractive index 1.5. what is the frequency and wavelength of the light in the glass block so they're asking frequency in the glass and also wavelength in glass you know the formula for refractive index i just taught you which is c by v okay you know mu is 1.5 see c is 3 into 10 to the power 8. I will find V with this. Which is velocity of light and velocity of dirt in the medium. Correct? Medium. Where velocity lambda is given. Which is in vacuum. I want only. I don't want vacuum. I don't want vacuum at all. Because if you do as you say, I will not be able to use wavelength in vacuum. Correct? So with this I can find the velocity in the medium. which is 3 into 10 to the power 8 by 1.5 which will become 2 into 10 to the power 8 meter per second i get the velocity in the medium and i told you what is the formula for velocity is wavelength into frequency wavelength into velocity is distance by time which is wavelength into frequency so you know velocity of the medium you you know frequency does not change because frequency in air and glass are the same have they given the frequency no they given only wavelength in vacuum with which you have to find the frequency in vacuum for which you need the formula which is nothing but what is the formula velocity is equal to mu is equal to c by lambda so if you mu is equal to c by lambda so you want wavelength frequency in vacuum it's 3 into 10 to the power 8 by 4500 into 10 to the power minus 10 with mu is equal to c by lambda puta enak ena kadicharudha frequency in vacuum adha abdey eduthru vanda you put it on ma because the medium frequency is not going to change so you can do you know the you know the velocity in the medium so fill a 2 into 10 to the power 8 is equal to lambda in the medium into frequency which is general frequency which you can substitute this here 3 into 10 to the power 8 by 4500 into 10 to the power minus 10 but i can get lambda that will give you if that's what you have to do how can i do this any other easy method strikes you please go ahead and do it there's no fixed way okay this is wavelength in medium what are they asking then the wavelength in of the light in the glass block frequency frequency does not change it's the same as what is in the air and I invariably found it already I already found out using the wavelength in vacuum I found out the frequency in vacuum using the formula mu is equal to C by lambda are you clear So, I will do example 16. So, example 16, a ray of light passes through a plane boundary separating two media whose refractive indices are, there is one medium which has got refractive index 3 by 2. There is another medium which has got refractive index 4 by 3. So, a ray is entering from a boundary of two media. So, one has got that refractive index, one has got, this is medium 1 and this will be medium 2. if a ray travels from medium 1 to medium 2 at an angle of incidence 30 degree so what an incident ray you have we need a immediately normal at the point of incidence and you know this angle is 30 degree okay 30 degree when it travels into angle of incidence 30 degree what is the angle of refraction What is the angle of refraction r? And you know mu is equal to sine i by sine r. But in the mu, it is the second medium with respect to first medium here. Which is nothing but mu of the second medium by mu of the first medium. Correct? Is equal to sine i by, sine i is what? Sine i by sine r. What is mu 2? 4 by 3. What is mu 1? 3 by 2. The light ray is entering from rarer into denser then this formula is correct. It is entering from rarer to denser, mu is equal to sin i by sin r is correct. So mu of 2 with respect to 1 is correct. So, 4 by 3 by 3 by 2 is equal to sin i which is sin 30 by sin r. You can find out cross middle. What is sin 30? Sin 30 is 1 by 2. So, you can just do the calculation and get the answer for sin r. Are you clear? So, then that is about it. Next part of the question. If the ray travels from medium 2 to medium 1. If the ray travels from medium 1 to medium 2. In the second case, it is going to travel from. medium 2 which is a denser medium to medium 1 which is a rarer medium. The light is entering here and the normal is at the point incidence when it's entering from density to rar it will move away from the normal this is your i this will be your r what is the refractive index of second medium we saw it is 4 by 3 mu 2 is 4 by 3 what is the refractive index of first medium we saw it to be 3 by 3 by 2 correct now since it's entering from density to rar i can't write mu is equal to sine wave center i have to write 1 by mu is equal to sine wave center when am i centering from dense So, it's 1 by mu is equal to sin i by sin r. Whenever it's entering from rar iterance, mu is equal to sin i by sin r. Clear? Teacher, don't get confused. Write it like how I'm saying. So, it's 1 by mu is equal to sin i by sin r. Which is always, for me, mu is always. 2 with respect to 1. That does not change. I don't change, switch that 1 and 2. I will always keep it as 2 with respect to 1 but instead I will write 1 by because 1 by mu of 2 with respect to 1 is nothing but mu of 1 with respect to 2. Instead of that, I will write 1 by. So that is easier for me. I do not want to zigzag confuse that 1 and 2. I will keep it as 2 with respect to 1. If I want to change the medium, then I will write 1 by. If it is denser to the error, it is 1 by mu is equal to sin i by sin r. Correct? What is the angle of incidence? What is the angle of refraction? Oh, so the angle of incidence here is same, 30 degree it is entering. They are asking for angle of refraction and you write 1 by mu. What is mu? Normally, you are rewrite mu2 by mu1. Here you have to write mu1 by mu2 because it is reciprocal. So it is actually 1 by mu2 by mu1. Therefore, it will become mu1 by mu2 is equal to sin i by sin r. What is mu1? 3 by 2. what is mu 2? 4 by 3, correct? Is equal to sin i, you know is 30 degree by sin r and you know 30 degree is nothing, sin 30 is 1 by 2, just calculate and get the value for it and you know in this r will be much greater whereas here r will be much lesser. You understood how to do it, okay so let's go to next question. Example 17 is what I'm going to do next. A rectangular glass slab rests at the bottom of the trough of water. So let's draw a trough of water. A rectangular glass slab is lying underneath. It is in water. It is full of water. It is in water. A ray of light incident on the water surface at an angle of 50 degree passes through water into glass. So it comes from above. It is entering in. And then it is going to get into water. Normal. So it is going to bend towards the normal because it is entering from rarer medium into a denser medium which is water. This is air. So it is bending towards the normal and goes and falls on the glass slab. Again in the glass slab you should have one more. Normal because this is the incident ray. Correct? And you know it is further going to bend towards the normal because this is glass. Even denser than water. So it is going to bend so much inside towards the normal. Are you clear? So this is how I construct the diagram. And they have given that this angle is 30 degree. Angle at which it is incident. 50 degree? Okay. The angle at which it is incident. Normally 50 is not given. 30 is given in school. Okay. 50 degree. Okay. A rectangular glass up rests on the bottom of the glass. The ray of light incident on the water surface at an angle of 50 degree passes through water into glass. Calculate the angle of refraction in glass which means the question is R in glass. Refraction in glass given that the refractive index of water is 4 by 3 and the refractive index of glass. is 3 by 2. We already know 1.33 and 1.5. So, you can also write it as 4 by 3 and 3 by 2. But they will give it. If they give it, no problem. Now, you have to find what is r. Okay, let us explore this. First, let us find this. This is 50 degree, right? So, we can find this r because here it is entering from add to r. So, what is the refractive index formula? Mu is equal to sin i by sin r, no 1 by mu because it is entering from air to air to denser. Mu is equal to sin i by sin r where it is water with respect to air. What is water's refractive index? 4 by 3 is equal to sin 50 by sin r. So, when I cross multiply, I will get this sin r. So, let the sin r be. some angle approximately 35.06 35 points I told you know they won't give 50 degrees they'll give you 30 degree so that that refractor angle you can find it will be 45 degree something known angles only it won't be unknown angles okay so you find this angle once you find this angle which is let's keep it as 35 degree for now Keep it as 35. If you keep it as 30, it will be good. Keep it as 35. Now, we know this as 35. Interior alternate angle, this will also be 35. Z angle. So, the angle of incidence in the glass slab is 35. And it is going to enter into R and you have to find this. Okay, here it is glass and above which there is water. So, you have to write the refractive index of Is it entering from rarer to denser? Water to glass? Water to glass? Yes, so mu of glass with respect to water because it is entering from water to glass, you have to write mu of glass with respect to water which is mu of glass by mu of water. What is mu of glass? What is 3 by 2, what is mu of water? 4 by 3, which is what is equal to sine i by sine r, which is which i is this i? This i, what's happening in the glass lab, correct? What is sine i there? Sine 35 by sine... r you don't know sign r but you know 3 by 2 4 by 3 sign 35 you can calculate are you clear so slowly you should get it slowly every refractive index you should write and here alone in this medium alone you should know it's entering from water to glass so the refractive index will be with glass with respect to water and it's entering still from rarer to denser so i don't have to put 1 by mu i can put mu of glass with respect to water which is mu of glass by mu of water which is nothing but Sin i by sin r. That sin i you should find out. You can't take this i. Slowly, slowly you should use that and find. Are you clear? So, can I go to next sum? So, with these concepts, this is the intro concept. If you are clear, you will be able to do some basic sums which are one mark or two mark questions. The next concept is the glass slab experiment. You have done this. Glass slab will be there. There will be an incident ray. I am taking the derivation. At the point of incidence, you will have a normal. Extend the normal like this. So, it is entering from air into glass. This is air and it is entering into glass. So, it will bend towards the normal. At this point of incidence also, you need one more normal. It is going to enter from glass into air. So, it is going to bend again away from the normal. So if this ray, there was no glass slab, would have proceeded like that. If there was no glass slab, this ray would have gone straight like that. Do you want me to do it once again? So you have a glass slab. there is an incident ray at the point of incidence you have to draw a normal light ray is going to bend towards the normal because it is entering from air into glass say just answer then at this point of incidence again you need to have a normal it is entering from glass into back again air which means it is going to bend away from the normal which means this ray the incident ray would have gone like this if there was no glass which means it is displaced like this this is lateral displacement D we pointed up the put the latterly displaces called lateral displacement lateral displacement are you clear through a glass lab okay so you know you have to mark I is the angle of incidence incident ray makes to the normal mark are this is the reflected ray refracted ray makes with the normal and interior alternate angle can we call this also as therefore are zigzag correct, z maari rukka, this will be angle of emergence correct, for change let's call this as, okay, I am going to put a red color cross mark, very very important, can you see this is a cross mark here, Let's mark this now as R1 and R2 for now. So let's keep this will be I minus this small angle in here will be I minus R1 the gap and that is enough for me. I don't want anything more because I am trying to find this D and I have this triangle here. the whole objective is to find the lateral displacement d and i want only sign i minus r which has got opposite as d and hypotenuse okay the thing is the glass lab thickness is given write this point as a write this point as b write this point as n write normally write as n a knows and b knows that's hypotenuse that is not t only an is t but still i can't use t because i only know the i only know this eddy and i can only use the hypotenuse. I can only use the hypotenuse because this is 90 degree. Are you clear? So, therefore, I can do sin I minus R is equal to opposite which is d by hypotenuse. But I do not know hypotenuse. So, I have to use the other triangle. And do cos R. Cos R will from this triangle will be the thickness of the glass slab T by the hypotenuse. Therefore hypotenuse will be T by cos R and I can substitute that here. So what will happen here? It will be D by cos R. is equal to sin i minus r. What is the purpose to find? D. Just cross multiplication. D therefore is T into sin i minus r by cos r. And you should memorize this formula. Lateral displacement D is sin i minus r by cos r. Are you clear? And the red color cross. That i is r. That i minus r. r1. So, if you take that right triangle, displacement will come. So, sin i minus r is equal to D by hypotenuse. But I don't know hypotenuse. Take the adjacent. triangle. Find cos r. So that you get the hypotenuse. So t in the formula t into sin i minus r by cos r is known as lateral displacement formula. You should know the derivation also. So now that you know this concept, I will teach you some in this concept which will be example 18. A parallel sided glass slab of thickness 6 cm is made of a material of refractive index root 3. When the light is incident on one of the parallel faces at an angle of 60 degree. normal varianoula at an angle of 60 degree. Refractive index of the material is root 3. Then when the light is incident on one of the parallel plates at an angle of 60 degree, it emerges from the other face. find the lateral shift lateral shift kick now so then if diagram complete bending towards the normal then again one more normal here correct and it's going to enter away from the normal and you know this is how it's going to be and they're asking not for you know this will be R and we will do a cross like this remember and it will be sine I minus this ray would have gone like this If there was no glass lamp, this is what is lateral shift. Again I am saying again. This cross mark is what is. This will be I minus R. So 60 minus R. That becomes 60 minus R. And you can write the formula of lateral shift. Now I am doing it. D is equal to T into sin I minus R by cos R. But this R is not known. If R is not known, it is entering from air to glass. Refractive index is known. With that mu is equal to sin I by sin R. So mu of glass is equal to. So, sin i by sin r which is sin 60 by sin r root 3. What is sin 60? Sin 60 is root 3 by 2. Superb. So root 3 by 2 means root 3 and root 3 will get cancelled. So we will get sin r is equal to 1 by 2. Where will r be 1 by 2? 30 degree. So r is equal to 30. So I will find out this r is 30 degree. So you know r and I. So now we will substitute. Thickness of the glass slab is 6 cm. 6 cm is 6 into 10 to the power minus 2 into sin i minus r is 60 minus 30 by cos r. Cos r is cos. 30 substitute the value get the answer how is it so if unless you know this formula which is the derivation of the previous one you can't do the sum so let's go to next question the next question is what are they asking find the lateral shifter lateral shift is what is d Yeah, I found it directly. That's all. Nothing. I don't know what he is writing. So let's go to next question. The velocity of light. 19th question I am doing. The velocity of light in glass which is velocity of light in glass is medium. 2 into 10 to the power 8 meter per second. And in air, you know it's 3 into 10 to the power 8 meter per second. They don't even have to give. You know it. By how much would an ink dot appear to be raised? For this I have to teach you a concept again. So I will stop this sum and will do the concept first. The concept is this. Take a beaker. Fill it with water. You have a mark here. You can see it from the side. When you see it from the side, it will come into water first. Then you have to draw a normal at the point of incidence. Then the light ray is going to travel from denser medium into a rarer medium. So it is obviously going to bend away from the normal. So it is going to appear to come from a point here. The object is appearing to be raised. The object is appearing to be. This is where your object is. This is where your image is. Just construct a straight line like this. Then mark this as this point of incidence. This is object. This is A. This is B. Let's name it A. it like this okay so then mark it as O, A and B This explains why swimming pool appears shallow. Many children think swimming pool is this deep. They will jump into the pool. That is why children keep the gate of swimming pool in their school. They will think there is only water till here. But if you leave your head and look like this, along normal, you will get to see the exact depth. Okay, so why is it we are able to see them because it is along the normal. There is angle of incidence is 0 and angle of refraction is also 0. So, you get to see the actual depth. Side of patina, it will appear to be raised. Okay, and the final formula in that is mu is equal to actual depth. by apparent depth. This is the final formula. mu is equal to actual depth by depth. How are we going to derive that? Angle of incidence is equal to i. Angle of refraction is equal to r. Interior alternate angle becomes i. Corresponding angle becomes r. One second. Is it recording? So, you know the final formula is mu is equal to axial depth by apparent depth. So, then this is I, this is R. Can you see Z angle? So, this is I, this is also I. Corresponding angle, this is R, this is also R. This is very important. The light rays here is entering from denser to rarer or rarer to denser. Eye. Denser to rarer. Because object is in the bottom of the tank. To reach the person's eye, it has to enter from, kandlein da rai poothu. Object lin da rai poothu, kanduke. So, it's entering from denser to rarer. So, you have to write 1 by mu is equal to sin i by sin r. I told you. What is it? 1 by mu is equal to sin i by sin r. For me, mu is always denser medium with regard to rarer medium. For me, denser with regard to rarer. So, when you write that, mu is equal to sin r. So, this is equal to sin i by sin r. Normally, it is mu 2 by mu 1. Here, it will be mu 1 by mu 2 is equal to sin i by sin r. Correct? Which is equal to refractive index. But what is sin i from this triangle? Can you take this triangle? What is sin i? Opposite by hypotenuse. Sin i is opposite by hypotenuse. What is opposite? a b. Hypotenuse is o a. So it's a b by o a. By what is sin r? What is sin r? Sin r is this small triangle, right? which is opposite which is AB by hypotenuse which is IA. So AB by IA. So AB and AB gets cancelled. So you will get IA by OA correct so mu is not there one by mu it's one by mu is equal to sine I by sine R sorry I should keep it as one by mu I keep it as one by mu is equal to sine I by sine R so it's IA by OA what is IA what is the meaning of IA it's apparent IA is hypotenuse. IA is the hypotenuse. I don't want hypotenuse. In physics you can take hypotenuse to be approximately equal to sin to be tan. Sin theta can be written as tan I. Sin R can be written as tan R. So, then it will be... Ia ke badla you can write the height, Ia ke badla you can write Ib and Oa ke badla you can write Ob. So Ib is the apparent depth, Ob is the actual depth, correct? So this is 1 by mu. Therefore mu itself is actual depth by apparent depth. Do you understand how I got this? Very very important derivation. If you know this derivation then you can do this sum. So let me do this sum now. The sum is the velocity of light. 19th question. In glass is 2 into 10 to the power 8 meter per second. And that in air is 3 into 10 to the power 8 meter per second. by how much would an ink dot appear to be raised? supposing there is an ink dot how much would it appear to be raised? when seen from the side, correct? by how much would it raise? it's covered by a glass plate oh this is water no it's not like that this is this is a different sum you have an ink dot on which you are placing a glass plate the glass of a crack or glass switch it to the Cuddy in the mobile Cuddy lower daughter primary imagine your mobile to be one glass plate prena so then there also it is going to be raised only because it will know it has to travel from glass into air and it's going to bend and turn to appear to be raised only definitely correct ah so by how much wood and ink dot appear to be raised which covered by glass of thickness glass plate of 6 centimeter thick which is 6 into 10 to the power minus 2 meter thickness okay and you know apparent what is it refractive index formula is actual depth by apparent depth you Do you know refractive index of glass? It is 1.5 or 3 by 2. So then therefore you can write 3 by 2 is equal to the actual depth will be obviously the thickness of the glass slab itself. So 6 into 10 to the power minus 2 meter by apparent depth. With this you can class multiply 3 and 2 will get cancel 2 2's are 4 apparent depth therefore it is 4 into 10 to the power minus 2 meter by apparent depth. Once apparent depth is obtained, apparent depth is the depth from above. Correct? But they will ask you, how much is it appearing to be raised? They ask in this distance. Appearing to be raised is, actual depth minus apparent depth. If the question is appearing to be raised, raised will be actual depth minus apparent depth. What is actual depth? 6 into 10 to the power minus 2. Minus what is apparent depth? 4 into 10 to the power minus 2. So, 2 into 10 to the power minus 2 meters is the, where it is appearing to be raised. They have written as 2 centimeter. It does not matter. You can write your own. Can I go to next question? Next question, let us do example 20. A glass beaker of height 10 centimeter.