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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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1.5R

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Detailed Solution

$\frac{μ_{2}}{OQ} - \frac{μ_{1}}{(- OP)} = \frac{μ_{2} - μ_{1}}{R} \Rightarrow \frac{1.5}{x} + \frac{1}{x} = \frac{1}{2 R} \Rightarrow x = 5 R$

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