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Question

A light ray is incident at an angle 45o with normal on a 10cm thick plane slab of refractive index n=1.5. The lateral shift of the incident ray is ______ cm.

A
0.835
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B
8.35
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C
0.03
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D
1.197
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Solution

The correct option is C 0.03
Given,t=10cm=.1msinisinr1=R.Isin450sinr1=1.5sinr1=12×101.5=0.4713r1=sin1(0.4713)=2807α.s=0.1sin(4528.7)cos2807=0.1×0.29070.8820=0.03m
1236380_1498164_ans_342da54ec5504592aa9061a9e7c333f4.JPG

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