AIEEE and Earlier JEE Main Questions: Refraction

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. You shine a ray of light on one face of a glass prism, and it comes out of the other face. Now you tilt the ray closer to the normal. Surely it gets through even more easily, right? Well, at some angle, the light suddenly stops coming out. So where does it go? Today we solve two past questions, from AIEEE and JEE Main, and the second one answers this puzzle. So, let's begin.

0:29Critical angle and total internal reflection

First, a quick recap. Snell's law says n one sine i equals n two sine r. n one and n two, with the small numbers below, are the refractive indices, and i and r are the angles of incidence and refraction, from the normal. Okay? Now the figure. The source, marked S, sits in the denser medium of index n one, with the rarer medium above. The h on the dashed vertical line, the r marked along the surface, and the second formula on the right belong to another problem, so ignore them today; that r is a radius, not an angle. Going from denser to rarer, r is bigger than i. So tell me, as i grows, how far can r go? It goes up to ninety degrees. Then the refracted ray just grazes the surface, like the ray marked grazing. That angle of incidence is the critical angle, theta c: theta, the Greek letter, with a small c below. Put r equal to ninety degrees in Snell's law, and sine theta c equals n two over n one, the same formula you see at the right of the figure. Right? For a medium in air, n two is one, so sine theta c is one over n. Both our questions today use this form. Got it? If i is greater than theta c, there is no refracted ray, and all the light reflects back, like the ray marked totally reflected. This is total internal reflection, TIR for short. With me? Now a quick question for you. Can total internal reflection happen when light goes from air into glass? No, it cannot, and that is a common trap! Going into a denser medium, r stays smaller than i. So TIR needs light going from denser to rarer, and i greater than theta c. Clear?

2:23The prism: where light can get stuck

Now, the prism, in the left figure. Capital A at the top is the prism angle. The ray enters at angle i, makes angles r one and r two with the dashed normals, and leaves at the angle e, the angle of emergence; the delta, the delta formula below the prism, and the graph on the right are about deviation, which we do not need today. The key geometry is r one plus r two equals capital A, always. Okay? So let me ask you this. The sum is fixed, so if r one gets smaller, what happens to r two? It gets bigger, yes, exactly! And at the second face, light goes from glass into air, denser to rarer. So total internal reflection is possible there; this is where light can get stuck. Right? So the ray comes out of the second face only if r two is less than or equal to theta c; that is the sign with the line under it. At exactly theta c, it just grazes out. Got it? And a topper's trick from the course: start every prism question by writing r one plus r two equals capital A, and mark which angles are known. With me?

3:34Example 1: ray grazing the wall of a glass rod, find theta

Now example one, from AIEEE two thousand nine. Here, a transparent solid cylindrical rod, of index two over root three, is surrounded by air; the sign over the three is the square root. Then a ray enters the centre of one end at angle of incidence theta, and grazes along the wall. Find theta. Pause the video and try it first. Done? Okay, let's check. The ray bends at the end, then grazes along the wall. So tell me, what is its angle of incidence at the wall? It is the critical angle, yes! Grazing along the wall happens exactly when the angle of incidence there equals theta c. Okay? The rod is in air, so sine theta c equals one over n, which is root three over two. The double arrow means which gives: theta c is sixty degrees, and the small circle means degrees. Right? Now many students write the refraction angle at the end as sixty degrees. Is that correct? No, no, careful, that is the trap! Sixty degrees is measured from the normal to the wall. The wall is perpendicular to the end face, that is the upside down T sign, so the two normals are at right angles, and r plus theta c makes ninety degrees. So r, the refraction angle at the end, is ninety minus sixty, which is thirty degrees. Clear? Now Snell's law at the end. Air has index one, so one times sine theta equals two over root three, times sine thirty degrees, which is one half. With me? So sine theta equals one over root three, and theta is sine inverse of one over root three; the minus one up top means inverse, not a power. Got it?

5:20Example 2: angles for which light leaves the prism's face AC

Now example two, a harder one, from JEE Main twenty fifteen. Monochromatic light, of one single colour, is incident at angle theta on the face A B of a glass prism. The prism angle is capital A, and the index is mu, the Greek letter, our n. For what values of theta is the ray transmitted through the face A C? Pause the video and try it first. Done? Okay, let's check. The light enters at the face A B, and must leave through the face A C to get out. So tell me, where can it get stuck: at A B, or at A C? Yes, exactly, it is at the face A C! At A B, light goes from air into glass, so no TIR there, but at A C it goes from glass to air. So r two must be less than theta c, which is sine inverse of one over mu. Equal would only graze, so we keep it strictly less. Okay? Now the trick: r one equals capital A minus r two. Since r two is less than theta c, r one is greater than capital A minus theta c. Right? Now let me ask you this. To make r one bigger, should theta be bigger or smaller? It should be bigger, absolutely! Snell's law at A B gives sine theta equals mu sine r one, and the sine keeps increasing up to ninety degrees. So sine theta is greater than mu times sine of capital A minus theta c. Clear? Now put theta c back in. Theta is greater than sine inverse of, in square brackets, mu times sine of capital A minus sine inverse of one over mu. That is just the line above, with theta c written out in full. Got it? And now, our puzzle. If theta is smaller than this, r two becomes bigger than theta c, and the light is totally reflected at the face A C, so nothing comes out. Going in closer to the normal made it harder to get out! With me? And this is the course result for the smallest angle of incidence, i min, short for minimum, with min written small below. So, sine i min equals n times sine of capital A minus theta c, our answer with n for mu. Okay?

7:42Quick revision: TIR, sin c = 1/n, r1 + r2 = A, r2 at most c

So, quick revision. Total internal reflection needs denser to rarer, and i greater than theta c; in air, sine theta c is one over n. Grazing at a surface means the angle of incidence there is theta c. And in a prism, r one plus r two equals capital A, and light comes out only if r two is at most theta c. Now try the practice problems on this page. All the best!