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Question

One end of a glass rod of refractive index n=1.5 is a spherical surface of radius of curvature R. The centre of the spherical surface lies inside the glass. A point object placed in air on the axis of the rod at the point P has its real image inside glass at the point Q (see fig.). A line joining the points P and Q cuts the surface at O such that OP=2OQ. The distance PO is.
876738_341e4bd634b14486a4588a89319680f3.png

A
8 R
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B
7 R
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C
2 R
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D
Non of these
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Solution

The correct option is A 8 R
For spherical surfaces, the following relation holds good
μ2vμ1u=μ2μ1R
where,
μ2=Refractive index of glass
μ1=Refractive index of air
v=OP
u= OQ
Given that OP=2OQ (in magnitude, and opposite in sign)
1.5OQ1OP=1.51R
1.5OQ12OQ=1.51R
4R=OQ
OP=2OQ=2(4R)
OP=8R

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