What Does the Ratio sin(φ1)/sin(θ4) Reveal About Light Refraction?

In summary: So the angle of the ray depends only on the contrast of the rate of horizontal advance of phase, and vertical advance of phase.
  • #1
Misr
385
0
[PLAIN]http://img843.imageshack.us/img843/6831/dsc00723mm.jpg
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?
 
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  • #2
Maybe you should compute it by yourself? I believe you are able to multiply fractions with proper cancellation.
 
  • #3
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
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
Maybe you should compute it by yourself? I believe you are able to multiply fractions with proper cancellation.
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!

Matimatically is okay since theta1=fi2 so their sin would be cancelled
so
sin(fi 1)/sin(theta 4)=n1\n5 right?
If you don't feel that multiplying fractions is sufficiently general to count as an "explanation"
That's it!
horizontal advance of phase
what do u mean by this?
 
  • #6
Misr said:
what do u mean by this?
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.
 

Related to What Does the Ratio sin(φ1)/sin(θ4) Reveal About Light Refraction?

1. What is the refractive index problem?

The refractive index problem refers to the challenge of accurately measuring and predicting the refractive index of a material. The refractive index is a measure of how much light is bent, or refracted, when it travels through a substance. This property is important in a variety of fields, including optics, materials science, and chemistry.

2. Why is the refractive index important?

The refractive index is important because it can tell us about the physical and chemical properties of a material, such as its density, composition, and molecular structure. It also plays a crucial role in determining how light interacts with a material, which is essential for many applications, such as lenses, prisms, and optical fibers.

3. How is the refractive index measured?

The refractive index can be measured using a variety of techniques, including refractometry, interferometry, and ellipsometry. These methods involve measuring the angle of refraction of a light beam passing through the material, or the phase shift of the light wave as it travels through the material. Each method has its own advantages and limitations, and the choice of technique depends on the properties of the material being studied.

4. What factors can affect the accuracy of refractive index measurements?

There are several factors that can affect the accuracy of refractive index measurements, including the wavelength of light used, the temperature and pressure of the material, and the presence of impurities or defects. It is important to carefully control and account for these factors in order to obtain reliable and precise measurements.

5. How is the refractive index problem being addressed?

The refractive index problem is being addressed through ongoing research and development in the fields of optics, materials science, and physics. This includes the development of new measurement techniques, as well as the use of advanced computational methods and simulations to better understand and predict the behavior of light in different materials. Collaboration between scientists from different disciplines is also crucial in tackling this complex problem.

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