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Q.

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

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a

3 R

b

2 R

c

5 R

d

1.5 R

answer is A.

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Detailed Solution

μ2v−μ1u=μ2−μ1R

∴1.5x−1−x=1.5−1R

⇒x=5R

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