The figure below shows the path of a beam of light through several layers (n 1 =

Question

The figure below shows the path of a beam of light through several layers (n1 = 1.59, n2 = 1.38, n3 = 1.18 and n4 = 1.00) of different indices of refraction.

If ?1 = 31.6o, what is the angle, ?2, of the emerging beam?
5.64

The figure below shows the path of a beam of light through several layers (n1 = 1.59, n2 = 1.38, n3 = 1.18 and n4 = 1.00) of different indices of refraction. If ?1 = 31.6 degree, what is the angle, ?2, of the emerging beam? 5.64 x 101 deg What must the incident angle, ?1, be in order to have total internal reflection at the surface between the n3 = 1.18 medium and the n4 = 1.00 medium?

Explanation / Answer

nsin theta1 = n2 sin theta 2

1.59 sin 31.6 = 1.38 sin t

t = 37.1 degrees

1.38 sin 37.1 = 1.18 sin t

t = 44.86 degrees

1.18 sin 44.86 = 1 sin t

t = 56.34 degrees

at critical ab gle the refracted ray is at 90 degrees to normal

1.18 sin t = 1 x sin 90

t = 57.93 degrees

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