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Question

Figure shows an irregular block of material of refractive index 2. A ray of light strikes the face AB as shown in the figure. After refraction it is incident on a spherical surface CD of radius of curvature 0.4 m and enters a medium of refractive index 1.514 to meet PQ at E. The distance OE is
6691.png
  1. 8m
  2. 6m
  3. 4m
  4. 12m

A
4m
B
8m
C
12m
D
6m
Solution
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By Snell's law
μ1sini=μ2sinr
From the question,
1×sin45=2×sinr
Solving this, we get r=30
For this r, the ray after refraction from the first surface will travel parallel to PQ as shown.
For refraction at curved surface CD, it will appear as though u= as the ray is parallel to principal axis PQ.
Given R=0.4m
Putting this and the given values in the formula

μ2vμ1u=μ2μ1R
we get
1.514v0=1.51420.4

Solving this we get v6m

339301_6691_ans.png

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