A spherical surface of curvature R separates air $(μ=1.5)$. 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
Answer & explanation
Correct answer: option 2
As $\frac{μ2}{v}-\frac{μ1}{u}=\frac{μ-1}{R}$
or $\frac{1.5}{x}-\frac{1}{x}=\frac{1.5-1}{R}⇒x=\frac{2.5R}{0.5}=5R$