# Angle Spread with Snell's Law

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1. Apr 3, 2016

### OhBoy

1. The problem statement, all variables and given/known data
Dispersion in a window pane. In Figure (a) below , a beam of white light is incident at angle θ1 = 67o on a common window pane (shown in cross section). For the pane's type of glass, the index of refraction for visible light ranges from 1.535 at the blue end of the spectrum to 1.518 at the red end. The two sides of the pane are parallel. What is the angular spread of the colors in the beam (a) when the light enters the pane and (b) when it emerges from the opposite side? (Hint: When you look at an object through a window pane, are the colors in the light from the object dispersed as shown in, say, Figure (b)?)

2. Relevant equations
n1 * sin(θ1) = n2*sin(θ2)

3. The attempt at a solution
Part a is straight forward. You use snells law from the red and blue end as it enters left side to get angle it leaves right side and take the difference to get the spread. It's part b I'm clueless about. We're not given any information on the triangle, and even if we had it I'm still confused because the light does not right the right side at the same point so I don't think we can just take the difference. I guess you could extend the light coming out of right side back into the triangle to treat it like its own triangle you need to get angle of, but wouldn't the angles that come back outward be dependent on the angle of head of triangle? Any ideas?

2. Apr 3, 2016

### Staff: Mentor

I think they're looking for you to arrive at some conclusion about the angular spread (divergence) for a flat pane of glass. The particulars of the prism don't matter, it's enough to note that the color trajectories diverge at the exit of the prism (hence the "rainbow effect" that prisms are noted for).

Sketch the trajectories for the two frequency extremes What's your conclusion?

3. Apr 3, 2016

### OhBoy

When I just use an equilateral triangle and plug in the values to get the angle from the normal of the right side the light leaves, I got 36.8 degrees for red and 37.124 degrees for the blue end.

I got the angle coming in on the right side by using the angle refracted on left side, subtracting 90 by that amount, then subtracting 180 by that amount and by 60, then subtracting 90 by previous amount. Then I used snells law.

4. Apr 3, 2016

### Staff: Mentor

As I said, you don't need to know any details about the prism scenario other than that it causes the colors to diverge. The amount of divergence will depend upon the geometry of the prism, but you don't care about the specifics.

Concentrate on the flat window pane. How are the colors arranged on exit from the window pane? Is there separation? Divergence?

5. Apr 3, 2016

### OhBoy

Why aren't the specifics relevant? It wants a specific angle spread.

The colors are arranged in order of low frequency to high frequency. It appears to diverge to a greater degree than inside the prism.

6. Apr 3, 2016

### Staff: Mentor

The question was...
...for the window pane ONLY. The prism was used in the hint as an example of when there is divergence (colors spreading out).

So work with the flat pane only to answer parts a and b.

7. Apr 3, 2016

### OhBoy

Ouch. For some reason I was assuming part a was telling us that the front and back sides are parallel. Thought I was only working with the prism and getting the spread from the right side of the prism :S. Sorry. Light enters at 67 degrees and should leave at 67 degrees so there is no spread.

8. Apr 3, 2016

9. Apr 3, 2016

### OhBoy

So the red light gets refracted by angle x, purple light by angle y, Angle y is less than angle x. We can get the change in vertical direction relative to when it hits left side by taking tan of angle x and multiplying it by the width. Then you can get the change vertical height of purple light by taking tan (y) and multiplying it by width, the separation would be the difference between the two? so W*(tan(x)-tan(y))?

10. Apr 3, 2016

### Staff: Mentor

Since no specifics were given about the width of the pane it should be enough to conclude that there will be a separation of the colors, but no dispersion. It's a form of chromatic aberration, where the images seen through the glass have slightly different locations at different wavelengths when viewed obliquely through a window.