Displacement of emerging light ray

In summary, 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.
  • #1
jssamp
31
3

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?
 
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  • #2
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"?
 
  • #3
sin(theta) = theta for theta < .2 radians
 
  • #4
jssamp said:
sin(theta) = theta for theta < .2 radians

Yes, anything else? (Hint: Another trig function has almost the same value for small angles)
 
  • #5
oh, yeah, cos(theta) = 1 - theta/2
 
  • #6
I might use that back when I used the sin(A-B)=sinAcosB-sinBcosA identity.
 
  • #7
I'll try it from the start with this added info. thanks gneill.
 
  • #8
There's yet ANOTHER trig function that's ~θ when θ is small. You may find it even more convenient...
 
  • #9
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:
 

1. What is the displacement of an emerging light ray?

The displacement of an emerging light ray refers to the change in position or direction of the light ray as it passes through a medium or interface.

2. Why does a light ray experience displacement?

A light ray experiences displacement due to the change in the refractive index of the medium it is passing through. This change causes the light ray to bend or change direction.

3. How is the displacement of a light ray calculated?

The displacement of a light ray can be calculated using Snell's law, which takes into account the angle of incidence and the refractive indices of the two mediums the light ray is passing through.

4. What factors affect the displacement of a light ray?

The displacement of a light ray can be affected by the angle of incidence, the refractive indices of the two mediums, and the wavelength of the light. Additionally, the surface roughness or curvature of the interface can also play a role in the displacement of a light ray.

5. How does the displacement of a light ray impact its path?

The displacement of a light ray can significantly impact its path, as it can cause the light ray to bend, refract, or reflect off of surfaces. This can ultimately affect the final destination or direction of the light ray.

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