Reflection and Refraction

  • Thread starter sghaussi
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  • #1
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Hello! I'm having difficulty solving this problem, I was wondering if you can help me solve it.

Light is incident in air at an angle on the upper surface of a transparent plate, the surfaces of the plate being plane and parallel to each other.

A ray of light is incident at an angle of 66.0 on one surface of a glass plate 2.40 cm thick with an index of refraction 1.80. The medium on either side of the plate is air. Find the lateral displacement between the incident and emergent rays.


I know I am supposed to use the formula: n_1sin(theta_1) = n_2sin(theta_2)
index of refraction of air = 1.00


However I don't know how to approach this problem. I'm not asking for the answer, just maybe a few steps on how to approach this type of problem? And will I need to use trig to find the distance between the incident and emergent rays?


Thank you,
Sahar
 

Answers and Replies

  • #2
Doc Al
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Draw yourself a picture of what's going on. When the light hits the glass, it's refracted at an angle given by Snell's law. Calculate the angle, then use a bit of trig to figure out where the light exits the glass (compared to the where it enters). Then figure out where the light would exit the glass if it weren't refracted. Compare.
 
  • #3
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And remember that Snell's Law calculates the angle with respect to the normal line. Do they want the angle between them or the distance between them?
 
  • #4
Doc Al
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You need to calculate the lateral displacement between the incident and emergent rays. Note that the lateral displacement is the perpendicular distance between the two rays. (Not just the distance between the exit points.)
 
  • #5
GCT
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Draw a proper ray diagram, you'll find two triangles with the same base, in terms of the first and second incident angles, which can be compared to give a value for s, displacement. Simply derive the equation for s, it'll be in terms of the thickness and angle 1 and 2 (first incident and emerging rays).
 

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