Refraction through a Prism

Transcript

0:00Introduction

Hello everyone! Let us start with a small experiment. Send a thin beam of light through a glass prism, and watch the bright spot it makes on the wall. Now slowly turn the prism, always in one direction. The spot moves, then stops, and then comes back, even though you are still turning the same way! So why does it turn back? By the end of this video, you will find the exact angle where it turns. Prism questions come up often in JEE, so let's begin.

0:32Two relations for every prism

Look at the left half of the figure. The grey triangle is a glass prism, and capital A, the angle at the top, is the prism angle. The orange ray enters at the angle of incidence, i, and leaves at the angle of emergence, e, both measured from the normals, the grey dashed lines. The graph on the right, we will use in a few minutes. Okay? Inside the glass, the ray makes angles r one and r two with the normals; the small one and two are subscripts. Now look below the prism: r one plus r two equals capital A, the prism angle. So let me ask you this: if you change the angle of incidence, does this sum change? No, it does not change at all, and that is the beauty of it! So r one plus r two equals A, for every ray, whatever the angle of incidence. Right? Now, the orange dashed lines extend the incoming and outgoing rays, and the angle between them is the deviation, delta, the Greek letter near the top. Delta equals i plus e, minus capital A, okay? So begin every prism question with r one plus r two equals A, and mark which angles you know. Clear?

1:45Minimum deviation: the symmetric ray

Now, the right half of the figure plots the deviation, delta, against i, the angle of incidence. As i grows, delta falls to a lowest value, then rises again. That lowest value is delta m, delta with a small m below it, the angle of minimum deviation. And that is our spot on the wall: it turns back at the bottom of this curve! Got it? So what is special about the ray at the bottom of the curve? Look just below the lowest point: the graph says i equals e. The ray goes in and comes out at the same angle, so it passes symmetrically. Now tell me, what are r one and r two for this ray? They are equal, yes, exactly! The same angle outside gives the same angle inside, so r one equals r two. And they add up to A, so each one is A over two. Here is a topper's trick: at minimum deviation, r is always A over two, so you never need to solve for it. With me? And inside, the ray runs parallel to the base, like the orange ray in the left figure, for an isosceles prism, one with two equal sides. And the formula next to the curve? We build it next. Okay?

2:58The prism formula

Now the formula. Put i equals e into the deviation relation. Delta m equals two i, minus A, and the double arrow means it follows that i equals, A plus delta m, over two. Right? Next, Snell's law at the first face, with the prism in air, and air has refractive index one: sine i equals n, the refractive index of the prism, times sine r one, and r one is A over two. So sine i equals n times sine of A over two. Got it? Put in i, solve for n, and n equals sine of half of, A plus delta m, over sine of half of A, okay? This is the prism formula, so write it down, and remember, it holds only at minimum deviation.

3:45Thin prisms, and a prism in a liquid

Now a special case, the thin prism: small A, and light falling nearly along the normal. Then delta equals, n minus one in brackets, times A, which is really simple, right? Now tell me: for a thin prism, does the deviation change when you change the angle of incidence? Hardly at all; as long as the light falls nearly along the normal, delta stays almost the same. But remember the conditions, small A and near-normal incidence, because they will matter soon. Clear? And for a prism in a liquid of refractive index n m, n with a small m below it, replace n by n over n m. So glass in water deviates light less than in air, because n over n m is closer to one. Okay?

4:33When can the light get out?

Now, does every ray come out? At the second face, light goes from glass, the denser medium, into air, the rarer one. So it emerges only if r two is less than or equal to theta c, theta with a small c below it, the critical angle. Otherwise, total internal reflection traps it. Got it? And r one reaches theta c only at grazing incidence, never more. So what happens if capital A, which is r one plus r two, is greater than two theta c? Then no ray at all can pass through, whatever the angle of incidence! Because r two is A minus r one, and that is always more than theta c, so the light is trapped at the second face. That is what this line says, with the greater than sign. Right? Now the largest deviation, delta max, with max written below the delta, comes at grazing incidence, i equals ninety degrees, where the small circle means degrees, or at grazing emergence. Both give the same value. With me? And any deviation other than delta m, where the slashed equals sign means not equal, comes from two angles of incidence, with i and e swapped. That is why the spot on the wall passes the same places again. Clear?

5:49Example 1: n of a 60 degree prism with minimum deviation 30

Now our first example. The angle of minimum deviation of a sixty degree prism is thirty degrees. Find its refractive index. Pause the video and try it first. Are you done? Okay, let's check. Capital A is sixty degrees, and delta m is thirty degrees. Right? Now, many students just add the two angles and write sine ninety degrees over sine sixty degrees. Is that correct? No, no, that is the trap, because both angles must be halved! Sixty plus thirty is ninety, and half of that is forty five degrees; half of sixty is thirty degrees. So n equals sine forty five degrees over sine thirty degrees. Okay? Sine forty five degrees is one over root two, where the slash means over and the tick is the square root sign. Sine thirty degrees is one over two. So n equals root two. Got it?

6:48Example 2: minimum deviation and i, 60 degree prism, n = 1.5

Now a slightly harder one. Find the minimum deviation of a sixty degree prism of index one point five, and the corresponding angle of incidence. Pause the video and try it first. Are you done? Okay, let's check. Some students say delta m equals, n minus one, times A, which is zero point five times sixty, so thirty degrees. Is that right? No, no, careful, that formula is only for a thin prism, and sixty degrees is not small! So use the prism formula: sine of half of, sixty plus delta m, equals one point five times sine thirty degrees, which is zero point seven five. With me? Now take sine inverse of zero point seven five, and sixty plus delta m, over two, equals forty eight point six degrees. Clear? Double it, and sixty plus delta m is ninety seven point two degrees. Then ninety seven point two minus sixty gives thirty seven point two degrees, not thirty. Okay? And the angle of incidence? At minimum deviation, i equals, A plus delta m, over two, the angle we just found, forty eight point six degrees. So for this prism, the spot on the wall turns back when the light enters at forty eight point six degrees. Got it?

8:13Quick revision: r1 + r2 = A, prism formula, thin prism rule

So, quick revision. First, for every ray, r one plus r two equals A, and delta equals i plus e, minus A, okay? Second, at minimum deviation, i equals e, r is A over two, and n equals sine of half of, A plus delta m, over sine of half of capital A, right? And third, delta equals, n minus one, times A, only for a thin prism, and light emerges only if r two is at most theta c, the critical angle. Now try the practice problems on this page. All the best!