Total Internal Reflection

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Put a small bulb at the bottom of a tank of water. Its light goes out in all directions, right? And yet only a round patch of the surface, right above the bulb, lets the light out. Why only a circle, and where does the rest of the light go? Today's idea, total internal reflection, answers both. It comes up often in JEE, so let's begin.

0:28Denser to rarer: the critical angle

Look at the figure. Below is the denser medium, with refractive index n one, that is n with a small subscript one. Above is the rarer medium, with index n two, smaller than n one. The dot marked S is the source, at depth h, and the dashed lines are normals. The letter r and the two formulas on the right, we will build ourselves in a few minutes. Now, a ray goes up into the rarer medium. Does it bend towards the normal, or away from it? Away from the normal, yes, exactly! See the ray on the left. So the angle of refraction is greater than i, the angle of incidence; that is the greater than sign on screen. Okay? Now keep increasing i, and the refracted ray bends further from the normal. At what angle of refraction does it just skim along the surface, like the ray marked grazing? Ninety degrees, yes, exactly! Then the refracted ray just grazes the surface. This angle of incidence is the critical angle, theta c, the Greek letter theta with a small c below it. The small circle after ninety means degrees. Right? Now apply Snell's law at the surface: n one sine theta c equals n two times sine ninety degrees. Sine ninety degrees is one, so sine theta c equals n two over n one. The double arrow on screen is read as, which gives. And this is the first formula on the right of the figure. Clear? If the rarer medium is air, n two is one, and n is the index of the medium. So in air, sine theta c equals one over n. Got it? Now push i beyond theta c. Then there is no refracted ray at all, and all the light is reflected back, like the ray marked totally reflected. This is total internal reflection. With me?

2:21Two conditions, and the values to remember

So, total internal reflection needs two conditions together. Light must go from denser to rarer, and the angle of incidence must be greater than the critical angle. Okay? Now here is a trap. Light goes from air into glass at a very large angle of incidence, say eighty five degrees. Can it be totally reflected? The answer is no, never. Air to glass is rarer to denser, so the first condition fails, and the ray bends towards the normal. Right? Some values from the course, for light going into air. Water, forty eight point six degrees. Glass of index one point five, forty one point eight degrees. And diamond, only twenty four point four degrees. Clear? Now, violet light has a larger refractive index than red light. So tell me, is its critical angle larger, or smaller? Smaller, that is correct! Sine theta c is one over n, so a larger n gives a smaller critical angle. Got it?

3:26A source under water lights a circle

Now, back to our bulb, the source S at depth h. Its rays hit the surface at all sorts of angles. Which of them can get out into the air? Yes, exactly! Only the rays whose angle of incidence, measured from the normal, is less than theta c get out; the rest are totally reflected. So only rays inside a cone of half-angle theta c escape, and that is why the bright patch is a circle! Its edge is where the grazing ray leaves, and its radius is the r marked across the top of the figure. With me? Now look at the right triangle with height h and base r. Here tan theta c is r over h. So r equals h times tan theta c, the second formula on the right of the figure. Right? In air, sine theta c is one over n. So in a right triangle with hypotenuse n and opposite side one, Pythagoras gives the adjacent side as the square root of, n squared minus one; that is the root sign on screen. So tan theta c is one over root of, n squared minus one. Okay? Put it in, and r equals h over root of, n squared minus one. Our second example uses this directly. Got it?

4:44Total reflecting prisms and optical fibres

Now let us see two uses. First, a glass prism with angles of forty five, ninety and forty five degrees. The ray enters a short face normally, so it goes straight in. It then meets the long face, the hypotenuse, at forty five degrees. What index does the glass need for total reflection there? We need forty five degrees to be greater than theta c. So one over n must be less than sine forty five degrees, which is one over root two. That means n greater than root two. Clear? Such a prism can turn a ray through ninety degrees, or send it back through one hundred and eighty degrees. Right? Second, the optical fibre: a core of index n one, inside a cladding of index n two, which is less than n one. Light in the core is totally reflected at the cladding, again and again. Okay? Light entering the end from air is trapped if its angle of incidence is below theta sub a, theta with a small letter a below it, where sine theta sub a equals the square root of, n one squared minus n two squared. Got it?

5:54Example 1: critical angle from glass into water

Now example one, an easier one from the course. Find the critical angle for light going from glass, of index three over two, into water, of index four over three. Pause the video and try it first. Done? Okay, let's check. Many students write sine theta c equals one over one point five, and get forty one point eight degrees. Is that right? No, and that is the trap! One over n works only when the rarer medium is air. Glass, at three over two, is denser, so it is n one, and water, at four over three, is n two. With me? So sine theta c equals n two over n one: four over three, divided by three over two, which gives eight over nine. Okay? So theta c is sine inverse of eight over nine; sine inverse, with the small minus one up top, means the angle whose sine is eight over nine. The wavy equals sign means approximately: about sixty two point seven degrees. Right?

7:01Example 2: smallest opaque disc over an underwater source

Now example two, a harder one, and it brings back our puzzle. There is a point source at the bottom of a tank of water forty centimetres deep, with index four over three. Find the radius of the smallest opaque disc that, floating on the surface above the source, stops all the light from coming out. Pause the video and try it first. Done? Okay, let's check. The disc only has to cover the bright circle, since outside it the light is totally reflected. So its radius is r equals h times tan theta c. Clear? Using our formula, r equals h over root of, n squared minus one. Here n squared is sixteen over nine, and sixteen over nine minus one is seven over nine. Got it? So r equals forty, over root of seven over nine. The root of nine is three, so this becomes one hundred and twenty over root seven centimetres. Okay? One hundred and twenty over root seven is about forty five point four centimetres. That is our bulb puzzle, solved with real numbers!

8:09Quick revision: sin c = n2/n1, 1/n in air, circle radius

Right? So, quick revision. Total internal reflection needs light from denser to rarer, and i greater than theta c. Sine theta c equals n two over n one, or one over n in air. And a source at depth h lights a circle of radius h times tan theta c, which is h over root of, n squared minus one. Now go and try the practice problems on this page. All the best!