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Prism Optics Range of Refraction Angles

  • Thread starter dvolpe
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1. Homework Statement
Equilateral triangle prism (angles all 60 degrees) made of crown glass. Index of refraction ranges from 1.486 for longest visible wavelengths to 1.556 for the shortest. What is range of refraction angles for the light transmitted into air through the right side of the prism?
Angle of white light entering prism is 55 degrees with respect to normal.


2. Homework Equations

n = [sin((60+D)/2)/(sin(60/2))]

3. The Attempt at a Solution
n min = 1.486 = sin ((60+D)/2)/[sin 30 deg)
1.486*.5 = sin (x)
x = (60+D)/2 = sin (.743)
(60+D)/2 = 47.987 or D = 35.974 degrees


This is WRONG! Help!
 

ehild

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Angle of white light entering prism is 55 degrees with respect to normal.


2. Homework Equations

n = [sin((60+D)/2)/(sin(60/2))]
Is this equation relevant? What does D mean? Why do you think the angle of incidence is given?

ehild
 
37
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I know that this equation can give me the D, the angle of minimum deviation, but I am not certain how that is important. I do not understand how to use the angle of incidence here.
 

ehild

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The minimum angle of deviation is a special angle of incidence, but the angle of incidence is given now.
You need the angle of refraction for both wavelengths. Apply Snell's law twice. Make a drawing, you will understand.

ehild
 
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I made a drawing and can get the first angle of refraction from Snells Law into the prism. Then I have trouble from that point on determining the angle of exit from the right side of the prism. Does the sin of angle of entry into prism times n for prism equal the sin of angle of exit on the right side times the n of the air?

When I do the above and apply Snell's Law I get that theta 2 into prism is 33.32 degrees. Then n of prism * sin 33.32 degrees equals n of prism times theta 3 (on right side before it exits) which means that theta 3 equals theta 2, or 33.32 degrees. Then n of prism times theta 3 equals n of air times theta 4 (refracting angle exiting prism into air), or theta 4 equals 55 degrees which is wrong! Please help.
 
Last edited:

ehild

Homework Helper
15,362
1,768
I made a drawing and can get the first angle of refraction from Snells Law into the prism. Then I have trouble from that point on determining the angle of exit from the right side of the prism. Does the sin of angle of entry into prism times n for prism equal the sin of angle of exit on the right side times the n of the air?
No.

Look at the picture. sin(α1)=n sin(β1). Calculate β1. Find out γ and then β2. Determine sin(α2)=nsin(β2).

ehild
 

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I understand most of what you are saying and drew a diagram. Would you reclarify something for me: Y is equal to B1 as they are alternating angles on the diagonal of the parallelogram. Then B2 equals x (opposite alternating angle on diagonal). How do I find x as I am not certain of the value of the apex of the upper triangle of the parallelogram containing x, y,and the apex angle. If I knew the apex angle, then x becomes 180-(apex +Y) and I then have B2.
 

ehild

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The yellow rectangle is not a parallelogram, but it has two right angles at A and B, and the third angle is 60° so

γ=360-90-90-60=120°.

In the triangle ABC, one angle is β1, you determine it from Snell's law. The other angle is γ, so you can find β2.


ehild
 

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