Question

In the given figure, the radius of curvature of curved surface for both the planoconvex and plano-concave lens is 10 cm and refractive index for both is 1.5. The location of the final image from the plano-concave lens after all the refractions through lenses is

• A. 15 cm
• B. 20 cm
• C. 25 cm
• D. 40 cm

Answer 1

The correct option is B

20 cm

The focal length of the plano-convex lens is

$\frac{1}{f}=(1.5-1)(\frac{1}{+10}-\frac{1}{\infty })=\frac{1}{20}$

Focal length of plano-concave lens is

$\frac{1}{f}=(1.5-1)(\frac{1}{\infty }-\frac{1}{10})=\frac{-1}{20}$

Since rays parallel first image formed +20cm from plane of convex lens. So for plane of concave u = +10.

$\therefore v=\frac{uf}{u+f}=\frac{10\times 20}{10-20}=20cm$