 Let's solve a few questions on critical angle and total internal reflection. For the first one, we have a torch light which stands upright under water as shown. A portion of the light emitted comes out of the water from a circular region on the surface and theta is the angle subtended by the circle, by this circle, on the light source. Our job is to find the angle theta. And we know that the refractive index of water is 1.2 and we need to report our answers in degrees up to 1 decimal place. As always, why don't you pause the video and first attempt this question on your own. Okay, so our job is to figure out this angle theta and all we know is the refractive index of water that is 1.2. Now in the diagram, we see that the portion of light emitted comes out of this water from the circular region. So we can imagine that there are all these light rays which are striking the surface and they are coming out from the surface of the water. There might also be some rays which are getting internally reflected and we can see that in a diagram as well, there are some light rays which undergo total internal reflection. So what we can do is we can try and draw how the light rays could look like. There can be light rays which strike the surface of the water and then they get refracted as they come out from the surface. There can also be some light rays which are undergoing total internal reflection and we know that light rays undergo total internal reflection when the angle of incidence exceeds the critical angle and the light rays undergo a refraction when the angle of incidence is less than the critical angle. So in between these two light rays, there will also be some rays which are incident at the critical angle and therefore the angle of refraction for them is just 90 degrees. They move along the surface of the water. For the first case, when the light rays are undergoing refraction and they are coming out of the surface, we do not really know anything. We do not know the angle of incidence. We do not know the angle of refraction. Similarly for the case when the light rays are undergoing total internal reflection, we still do not know anything. The angle of incidence could be anything. We know that it will be more than critical angle but that's just about it. We do not know anything else. But for the case when the light rays are incident at the critical angle for these black light rays, at least we know that the angle of refraction in this case will be 90 degrees and if we draw that it will look somewhat like this. The angle of incidence in this case is the critical angle and therefore the angle of refraction is 90 degrees. Now over here using SNES law, we can try and figure out the critical angle and let's try and relate the critical angle with this angle theta. We will come to that but first let's try and figure out the critical angle. So using SNES law, using SNES law this would be refractive index of the incident medium into the sign of the incident angle equals to the refractive index of the refractive medium into the sign of the angle of refraction. Here the incident medium is water and the refractive medium is air and we know that the refractive index of water is 1.2. So we can write 1.2 into sine of theta C. This is equal to refractive index of air that is 1 into sine of 90 degrees because that is the angle of refraction that is 90 degrees and we know that sine 90 is also 1. So sine theta C, sine theta C becomes equal to 1 divided by 1.2 and theta C would be sine inverse of sine inverse of 1 divided by 1.2 and this would come out to be equal to 56, 56.4 degrees. Okay, so now we know what the critical angle is and but we need to figure out this angle theta. Now if we had a normal that goes right through the source the torch light then these two normals, this normal right here and the normal that goes to the torch light they would be parallel to each other. So maybe then we should be able to figure out this half angle that would be equal to the critical angle because they will be alternate angles. So that would somewhat look like this. We have this angle as theta C because these two are alternate angles and this normal right here and this normal they are parallel to each other. Therefore you get alternate angles. So theta, theta basically comes out to be equal to 2 into theta C and this is equal to 112.8 degrees. So this right here is 112.8. All right, let's move on to a next question now. For the second one, we have two light rays which are emerging in a medium A and ray 2, ray 2 this one, it grazes the interface of A and B after striking it at an angle of 60 degrees. So this angle right here is 60. Ray 1 refracts into medium B and we can see that and then reflects at the interface of B and C. Okay, we can see that as well. No light is transmitted into medium C. Which of these are correct relations among the refractive indices of A, B and C and the refractive indices of A, B and C are N, A, N, B and N, C respectively. And here are your options. So we need to choose two answers out of these four. Again pause the video and give this one a try. All right, hopefully we have given this a shot. Now over here, we can see how the light rays behave when they strike the interface of any two, any two surfaces. So from this observation, we can try and reduce how the refractive indices of these three mediums are related. One thing that we can see between C and B is that the light ray in the medium B is undergoing total internal reflection at the interface of C and B and that is only possible when the light rays goes from a denser medium to a rarer medium. So the refractive index of medium C, it has to be less than the refractive index of medium B. Similarly, between B and A, we can see that the light ray is undergoing refraction and it is moving away from the normal. So refractive index of B, it has to be less than the refractive index of A. This is only possible when the light rays move from a more dense medium to a less dense medium. Now we can combine these two and we can write, we can write them as the refractive index of C, which is less than the refractive index of B and that is less than the refractive index of A. So just from this, we can have a look at the options and we can see that option B does not make sense. Option B is wrong because NB has to be greater than NC. Option C is correct. And also we can see that option D does not make sense because NA is greater than NB. But option D says that NB, this implies that NB is more than NA because you have to add two NAs to get one big NB. So this is wrong because refractive index of A, medium A, it has to be more than the refractive index of medium B. So this is also wrong. Now since the question says choose two answers, we can just pick the first one and say that is correct. But let's, let's arrive at the first option and see if it makes sense. For this one, we can try and look at the rate two because we have exhausted all the possibilities of ray one. Now let's look at ray two. For ray two, we can see that the angle of incidence is 60 degrees at the interface of A and B and the ray is moving along the interface. The angle of refraction is 90 degrees. Now using Snell's law for this ray, we can write refractive index of the incident medium into the sign of incident angle. Equals to the refractive index of the refracted medium into the sign of angle of refraction. Now Ni over here is just Na, that is the incident mediums refractive index into sign of 60 degrees. That is the incident angle. It is 60. This is equal to NB into sign of 90 degrees, which is sign of 90 degrees, which is one. Sign 60 degrees is root three by two. This is root three by two. So Na, the refractive index of medium A, this is equal to two divided by root three into the refractive index of medium B and two divided by root three is 1.2. So turns out option A is correct. Na is equal to 1.2 into the refractive index of medium B. All right, you can try more questions from the exercise in this lesson and if you are watching on YouTube, do check out the exercise link which is added in the description.