Question

# A spherical convex surface of radius of curvature R separates air (μa=1) from glass(aμg=1.5) .The centre of curvature is in the glass is O. 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
1.5 R
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Solution

## The correct option is A 5RUsing the relation μ2v−μ1u=μ2−μ1Rwhere u and v is the object distance and image distance, respectively.Given : μ1=1 and μ2=1.5Let the distance of object from point O be x.⟹ u=−x and v=x ∴ 1.5x−μ1(−x)=1.5−1R32x+1x=12R3+22x=12R⟹x=5R

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