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Question

# A spherical surface of radius of curvature R, separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O and PO = OQ. The distance PO is equal to :

A
5R
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B
3R
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C
2R
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D
1.5R
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Solution

## The correct option is A 5RGiven, PO=OQ⇒ u=vWe knowμ2v−μ1u=μ2−μ1R⇒μ2v−μ1−u=μ2−μ1R⇒μ2+μ1V=μ2−μ1R⇒1.5+1v=1.5−1R⇒2.5v=0.5R⇒v=2.5R0.5v=5ROP=OQ=u=v=5R

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