How does the angle of refraction affect the width of a transmitted beam?

AI Thread Summary
The discussion focuses on how the angle of refraction influences the width of a transmitted beam when it crosses an interface with different indices of refraction. It is established that the width of the transmitted beam remains the same as the incident beam when the interface is perpendicular to the incoming rays. However, if the interface is not perpendicular, the width can change depending on the angle of incidence. The relationship between the angles is governed by Snell's law, which states n1 sin(theta1) = n2 sin(theta2). Ultimately, the width of the exit beam can be expressed as a function of the incident angle, emphasizing the need for a general equation to describe this relationship.
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Homework Statement



(Refraction)
A parallel beam of width W is incident on a interface. The index of refractions are n1 on
the incident side and n2 on the transmitted side.
(a) Find an equation that gives the width of the transmitted beam. (Assume there is an
exit beam. IE we are not above the angle of total internal reflection.)

Homework Equations



n1 sin theta1 = n2 sin theta2

The Attempt at a Solution



If I draw a diagram, it seems that the refracted width is the same as the incident width. Anyone care to explain?
 
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haven't actually tried calculating the width of the exit beam?
 
yes, and the width is unchanged (?). both top and bottom "rays" of the beam are refracted at the same angle and come out of the interface parallel to each other.
 
Erm, unless the interface is not perpendicular to the incident beam...
 
yes, the question is asking for the general expression of the exit beam width in relation to the incident beam as a function of the incident angle (well, it wouldn't be wrong to write them as a function of the angle of refraction but it's trivial)
 

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