Displacement of emerging light ray

jssamp
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Homework Statement



A light ray is incident on a flat piece of glass with index of refraction n. Show that if the incident angle θ is small, the emerging ray is displaced a distance d = tθ(n - 1)/n form the incident ray, where t is the thickness of the glass and θ is in radians.

Homework Equations



n1sinθ1 = n2sinθ2

The Attempt at a Solution



Using Snell's law, right triangles, sin(a-b) identity, and small angle approximation, I have got it to:

d = tθ(n-(cosθ/cosθR))/n [θR is angle of refraction in glass]

how do I get cosθ/cosθR = 1?
 
Last edited:
jssamp said:

Homework Statement



A light ray is incident on a flat piece of glass with index of refraction n. Show that if the incident angle θ is small, the emerging ray is displaced a distance d = tθ(n - 1)/n form the incident ray, where t is the thickness of the glass and θ is in radians.

Homework Equations



n1sinθ1 = n2sinθ2

The Attempt at a Solution



Using Snell's law, right triangles, sin(a-b) identity, and small angle approximation, I have got it to:

d = tθ(n-(cosθ/cosθR))/n [θR is angle of refraction in glass]

how do I get cosθ/cosθR = 1?

You should be able to get to the result in a more direct way. What is your understanding of "the small angle approximation"?
 
sin(theta) = theta for theta < .2 radians
 
jssamp said:
sin(theta) = theta for theta < .2 radians

Yes, anything else? (Hint: Another trig function has almost the same value for small angles)
 
oh, yeah, cos(theta) = 1 - theta/2
 
I might use that back when I used the sin(A-B)=sinAcosB-sinBcosA identity.
 
I'll try it from the start with this added info. thanks gneill.
 
There's yet ANOTHER trig function that's ~θ when θ is small. You may find it even more convenient...
 
OK, I got it solved! Thanks for the help, I'd never heard about small angle approximation for cos and tan until this problem. The tan approximation was the key piece I was missing.

Thanks for the help!
 
  • #10
Glad to help :smile:
 

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