Question

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

• A. 5 R
• B. 3 R
• C. 2 R
• D. 1.5 R

Answer 1

Answer: (A)

Let us say PO = OQ = X
Applying$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_1-{\mu }_2}{R}$
Substituting the values with sign
$\frac{1.5}{+X}-\frac{1.0}{-X}=\frac{1.5-1.0}{+R}$
(Distances are measured from O and are taken as positive in
the direction of ray of light)
$\therefore \frac{2.5}{X}=\frac{0.5}{R}$
$\therefore X=5R$