Question

A plano-convex lens of focal length 20 cm has its plane side silvered:

• A. The radius of curvature of curved surface of plano-convex lens is equal to half of radius of curvature of a surface of equi-convex lens of focal length 20 cm
• B. A object placed at15 cm on the axis on the convex side of silvered plano-convex lens gives rise to an image at a distance of 30 cm from it
• C. An object placed at a distance of 20 cm on the axis on the convex side of silvered plano-convex lens gives rise to an image at 40 cm from it
• D. Silvered plano-convex lens acts as a concave mirror of focal length 10 cm

Answer 1

Answer: (A), (B), (D)

The correct options are
A The radius of curvature of curved surface of plano-convex lens is equal to half of radius of curvature of a surface of equi-convex lens of focal length 20 cm
B A object placed at15 cm on the axis on the convex side of silvered plano-convex lens gives rise to an image at a distance of 30 cm from it
D Silvered plano-convex lens acts as a concave mirror of focal length 10 cm

For plano-convex lens $\frac{1}{f}=\frac{(\mu -1)}{R}$
For equi-convex lens $\frac{1}{f}=(\mu -1)\frac{2}{R^{\prime }}$

$R=\frac{R^{\prime }}{2}$ (choice A is correct)
Let F be the focal length of silvered plano-convex lens
$\frac{1}{F}=\frac{2(\mu -1)}{R}+\frac{1}{\infty }=\frac{2(\mu -1)}{R}$
Given $\frac{1}{f}=\frac{1}{20}=\frac{(\mu -1)}{R}$
F = 10cm (it acts as concave mirror) choice (D) is correct
$\frac{1}{u}+\frac{1}{v}=\frac{1}{F}$
If u = 15cm, $\frac{1}{v}=\frac{1}{10}-\frac{1}{15}=\frac{3-2}{30}=\frac{1}{30}$
$\Rightarrow v=30$
If u = 20cm, $\frac{1}{v}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}\Rightarrow v=20$ (choice B is correct).