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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3 R

2 R

5 R

1.5 R

answer is A.

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Detailed Solution

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$

$∴ \frac{1.5}{x} - \frac{1}{- x} = \frac{1.5 - 1}{R}$

$\Rightarrow x = 5 R$

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