Question

The figure below shows the path of a light beam through several slabs with different indices of refraction. (n4 = 1.06)

The figure below shows the path of a light beam through several slabs with different indices of refraction. (n4 1.06) n1.60 1.40 1.20 (a) If 01 33.0°, what is the angle 02 of the emerging beam? (b) What must the incident angle 01 be to have total internal reflection at the surface between the medium with n 1.20 and the medium with n4 1.06?

Answer 1

Solution)

a) According to snell's law ,

sin 33 / sin θ2 = n4 / 1 .6 = 1 / 1.6

θ2 = 60.62 degree (Ans)

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b ) since total internal reflection occurs ,

θc in n = 1.2 ,

sin θc = 1 / 1.2 = 0.8333 ;

in n= 1.4 , sin i / sin θc = 1.2 / 1.4

sin i = 0.8333 * ( 1 .2 / 1.4 ) = 0.714

in n = 1.06 , sin θ1 / sin i = 1.4 / 1.06

or sin θ1 = 0.714 * ( 1.4 / 1.6)

so incident angle , θ1 = 38.66 degrees (Ans)

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Good luck!:)