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.

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 RFor spherical surfaces, the following relation holds goodμ2v−μ1u=μ2−μ1Rwhere, μ2=Refractive index of glass μ1=Refractive index of airv=OPu= OQGiven that OP=2OQ (in magnitude, and opposite in sign)1.5OQ−1OP=1.5−1R1.5OQ−1−2OQ=1.5−1R4R=OQOP=2OQ=2(4R)⇒OP=8R

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