Question

A piano-convex lens of refractive index 1.5 and radius of curvature 30 cm is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of the size of the object ?

Answer 1

Given,

${\mu }_L=1.5$

$R_m=30cm$, radius of curvature of mirror

The net power of the system,

$P=2P_L+P_m$ (two times refraction and one time reflection)

$\frac{1}{f}=\frac{2}{f_L}+\frac{1}{f_m}$. . . . . .(1)

The focal length of lens, is given by

$f_L=(\frac{{\mu }_L}{{\mu }_a}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

$f_L=(\frac{1.5}{1}-1)(\frac{1}{infinite}-\frac{1}{-30})$

$f_L=\frac{0.5}{30}=60cm$

The focal length of mirror,

$f_m=\frac{R_m}{2}=\frac{30}{2}=15cm$

Equation (1),

$\frac{1}{f}=\frac{2}{60}+\frac{1}{15}$

$f=10cm$

For the real image and the same size of the object,

Magnification, $m=\frac{v}{u}=1$

$v=u$

When object is at the center of the curvature of the system.

$R=2f=2\times 10=20cm$

The correct option is A.