Snell's Law and Refractive Index

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Picture a beam of red light going from air into water, where its wavelength becomes shorter. So a student argues that, in the water, the beam must now look bluish. Is the student right? Hmm, hold on to your answer. Today we learn the refractive index and Snell's law, and before this lesson ends, we will settle this puzzle. Refraction comes up often in JEE, and it all stands on these two ideas. So, let's begin.

0:33Refractive index: speed, wavelength, frequency

So, what is the refractive index of a medium? n equals c over v, where c is the speed of light in vacuum, and v is its speed in the medium. So n tells you how many times faster light travels in vacuum than in that medium. Okay? Now let me ask you this. When light goes from air into water, which of these change: its speed, its wavelength, or its frequency? The frequency does not change at all, and that is the key point. It is fixed by the source, so it stays the same from one medium to another. Right? But the speed and the wavelength both change, in the same ratio: v equals c over n, and lambda equals lambda nought over n. Lambda, the Greek letter, is the wavelength, and lambda nought, with the small zero below, is the wavelength in vacuum. Clear? And now, our puzzle. The colour we see depends on the frequency, not on the wavelength. The frequency did not change, so the red beam in water still looks red, and the student was wrong. Got it? Finally, the relative index. n two one is the index of medium two relative to medium one; the small numbers below are just labels. It equals n two over n one, which is also v one over v two, the speeds the other way round. With me?

1:55Snell's law: how the ray bends

Now the figure. The right half shows apparent depth, our next lesson, so look at the left half. The ray comes through medium one, of index n one, into the shaded medium two, of index n two. The dashed line is the normal. The first law says that the incident ray, the refracted ray and the normal all lie in one plane. Okay? Small i is the angle of incidence, and small r is the angle of refraction, both measured from the normal. The second law is Snell's law: n one sine i equals n two sine r. Write it down, because most refraction problems start here. Right? Now suppose medium two is denser, so n two is bigger than n one. Is r smaller or bigger than i? Which way does the ray bend? It bends towards the normal, yes, exactly! n two is bigger, so sine r must be smaller than sine i, and r is smaller than i, just as in the figure. Clear? And entering a rarer medium, the ray bends away from the normal, so r is bigger than i. Got it? So how much does the ray turn? The deviation is mod i minus r; the two straight bars mean the modulus, the size of the difference without its sign. With me?

3:13Parallel layers and optical path

Now, a stack of parallel layers. Snell's law applies at every surface, so n sine theta, with theta the Greek letter for the angle with the normal, has the same value in every layer. Clear? Written out, n one sine theta one equals n two sine theta two, equals n three sine theta three; the three dots mean and so on. Right? So let me ask you this. With air before the stack and air after it, at what angle does the ray come out? At the same angle at which it went in, that is correct! The direction in the last medium depends only on the first and last indices, not on the layers in between. Okay? Next, if light crosses a length t, in a medium of index n, its optical path is n times t. Got it? And the time taken? Light moves at c over n in the medium, so the time is t over that speed, which works out to n times t, over c. With me?

4:14Example 1: wavelength and frequency of 589 nm light in water

Now example one, an easy one. Yellow light of wavelength five hundred and eighty nine nanometres, the n m on screen, in air, enters water of index one point three three. Find its wavelength and frequency in water. Pause the video and try it first. Done? Okay, let's check. Lambda equals lambda nought over n, so five hundred and eighty nine over one point three three gives about four hundred and forty three nanometres. Okay? Now the frequency. The wavelength fell by a factor of one point three three, so the frequency falls by one point three three too, right? No, no, careful, that is the trap! The frequency is fixed by the source, so f, the frequency, equals c over lambda nought: three times ten to the power eight, over five hundred and eighty nine times ten to the power minus nine. The cross means times, and ten to the power minus nine turns nanometres into metres. With me? That gives five point zero nine times ten to the power fourteen hertz, the H z on screen, in air and in water. So it still looks yellow, just like our red beam stayed red. Clear?

5:30Example 2: angle of incidence twice the angle of refraction

Now example two, a harder one. Here, a ray in air falls on glass of index one point six, at an angle of incidence that is twice the angle of refraction. Find the angle of incidence. Pause the video and try it first. Done? Okay, let's check. Air has index one, so Snell's law gives one times sine i equals one point six sine r. Put i equal to two r, and we get sine two r equals one point six sine r. With me? Now, many students cancel the sine on both sides, and write two r equals one point six r. Can we really do that? No, no, wait, that is the trap! Sine is a function, not a number, so you cannot cancel it. Use the double angle formula instead: sine two r equals two sine r cos r. Okay? The ray is not along the normal, so sine r is not zero; that is the crossed equals sign. Cancel it, and two cos r equals one point six, so cos r equals zero point eight. So r is cos inverse of zero point eight, where the minus one up top means inverse, not a power. That is about thirty six point eight seven degrees; the wavy sign means approximately, and the small circle means degrees. Right? So the angle of incidence is two times that, about seventy three point seven degrees. Got it? Let us check it. Cos r is zero point eight, so sine r is zero point six. Then sine i equals two times zero point six times zero point eight, which gives zero point nine six, and one point six times zero point six is also zero point nine six. Snell's law holds, so our answer is right. Clear?

7:19Quick revision: n = c/v, Snell's law, n sin theta constant

So, quick revision. n is c over v; the frequency stays fixed, while the speed and the wavelength fall by the factor n. Snell's law: n one sine i equals n two sine r, and entering a denser medium, the ray bends towards the normal. Through parallel layers, n sine theta is constant; the optical path is n times t, and the time is n times t, over c. Now try the practice problems on this page. All the best!