Question

A biconcave lens is made of glass with refractive index 1.5 and has radii of curvature 20cm and 30cm if the 20cm surface is silvered then effective focal length of mirror formed is:

Answer 1

we know, If one surface of biconvex lens is silvered then,

effective power of system = 2 × power of lens + power of mirror

or, $-\frac{1}{f_{eq}}=\frac{2}{f_{l}}-\frac{1}{f_m}$

$R_1=+30cm,R_2=-20cm$

so, $f_m=\frac{R_2}{2}=-10cm$

$f_l=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

= (1.5 - 1)[1/20 - 1/(-30)]

= 0.5 × 50/600

= 0.5/12 = 1/24

so, $f_l=24cm$

now, $-\frac{1}{f_{eq}}=\frac{2}{24}-\frac{1}{-10}$

= 2/24 + 1/10

= (20 + 24)/240

= 44/240

= 11/60

so, $f_{eq}=-60/11cm$

hence, eeffective focal length of mirror formed is -60/11 cm