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A spherical surface of radius of curvature R, Separate air (RI,1.0) from glass (RI,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 'O' and PO=OQ. The distance PO is equal to

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1.5R

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Detailed Solution

Apply at curved surface $\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$

$\begin{array}{l} \Rightarrow \frac{1.5}{x} + \frac{1}{x} = \frac{1.5 - 1}{R} \\ \Rightarrow \frac{2.5}{x} = \frac{0.5}{R} \\ \Rightarrow x = 5 R \end{array}$

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