# Refractive index problem

1. Sep 7, 2011

### Misr

[PLAIN]http://img843.imageshack.us/img843/6831/dsc00723mm.jpg [Broken]
a light ray falls on parallel layers made of transparent materials which have different refractive indicies
so ,what does the ratio sin(fi 1)/sin(theta 4) depend on???
1-n1,n2
2-n2,n3.n4
3-n1,n2,n3,n4,n5
4-n1,n2
could you explain the right answer to me?

Last edited by a moderator: May 5, 2017
2. Sep 7, 2011

### xts

Maybe you should compute it by yourself? I believe you are able to multiply fractions with proper cancellation.

3. Sep 7, 2011

### sophiecentaur

Get stuck into the thing, using the definition of refractive index and the successive angles on the way through. If you are careful, you will find it gives you a very much simpler answer than you might imagine!

4. Sep 7, 2011

### Ken G

What's more, the simple answer is not one of the ones given! (Since 1 and 4 are repeated, I'm guessing 4 was supposed to be n1, n5). If you don't feel that multiplying fractions is sufficiently general to count as an "explanation", just note that the horizontal advance of phase is the same in all 5 media, as that does not change across any of the boundaries (or even with a continuous variation of n). Thus the angle of the ray depends only only on the contrast of the rate of horizontal advance of phase, and vertical advance of phase. But the rate of advance of phase along the direction of the ray depends only on the local value of n, so the angle of the ray is always controlled entirely by its local n value. Thus any contrast between original and final directions must be dependent only on...

5. Sep 7, 2011

### Misr

Matimatically is okay since theta1=fi2 so their sin would be cancelled
so
sin(fi 1)/sin(theta 4)=n1\n5 right?
That's it!
what do u mean by this?

6. Sep 7, 2011

### Ken G

Think of the wave as rows of wave fronts, like crests of water waves in straight lines. Now imagine one of the interfaces between the different media, and imagine how one of those wave fronts (or wave crests) would move along that horizontal surface (ignore all else, just think about how that wave crest travels along that interface). The wave crest is a point of fixed phase, call it phase=0 at the crest if you like, and that point of phase=0 moves along the interface at some speed. It doesn't matter what that speed is right now, what matters is that it is the same speed for the wave on either side of the interface-- it's the same speed for the incoming wave in n1, for example, as for the outgoing wave in n2, because it's just moving along the interface between n1 and n2. Now there is also a speed that the wave front itself moves, diagonally, through n1 and through n2, but that speed is different in the two media (it scales like 1/n). That is the speed that the phase=0 advances along the direction that the wave is actually moving, but how could it be different in the two media, yet have the same horizontal speed component (along the interface and everywhere else too)? The same way any varying speed can have the same horizontal component-- by changing its direction! That is the "reason" that the wave changes direction when n changes, which is called refraction.