White light propagating in air is incident at 45 degrees on the equilateral pris

Question

White light propagating in air is incident at 45 degrees on the equilateral prism of the figure.

http://session.masteringphysics.com/problemAsset/1036318/4/RW-30-28.jpg

Find the angular dispersion "gamma" of the outgoing beam if the prism has refractive indices nred=1.582 and nviolet= 1.633

ANSWER IN GAMMA DEGREES

I used this expression:


n(air)sin(45) = n(red)*sin(theta(r1)) and n(air)sin(45) = n(violet)sin(theta(v))

Explanation / Answer

n(air)sin(45) = n(red)*sin(theta(r1)) =>1x sin(45) = 1.582 sin(theta 1) => theta 1 = 26.549 degrees since it is a equilateral prism => theta 1 + theta 2 = 60 degrees => theta 2 = 33.45 degrees and 1.582 sin(theta 2) = 1 x sin(gamma 1) => gamma 1 =60.695 degrees similarly for violet n(air)sin(45) = n(violet)*sin(theta(v1)) => 1 x sin(45) = 1.633 sin(theta(v1)) => theta(v1)= 25.658 degrees => theta (v2)= 34.34 degrees => 1.633 x sin(theta(v2)) = 1 x sin(gamma 2) => gamma 2 = 67.1 degrees => gamma = deviation of both of the refracted rays = 6.408 degrees

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