Question

A convex lens of radii of curvature 20 cm and 30 cm respectively. It is silvered at the surface which has smaller radius of curvature. Then it will behave as $({\mu }_g=1.5)$

• A. concave mirror with equivalent focal length $\frac{30}{11}cm$
• B. concave mirror with equivalent focal length $\frac{60}{11}cm$
• C. convexmirror with equivalent focal length $\frac{30}{11}cm$
• D. convex mirror with equivalent focal length $\frac{60}{11}cm$

Answer 1

Answer: (B)

Focal length for lens,
$\frac{1}{f_L}=\left({\mu }_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
Here, $\mu =1.5,R_1=20cm,R_2=-20cm$
$\therefore \frac{1}{f_L}=\left(1.5-1\right)\left(\frac{1}{20}-\frac{1}{-30}\right)$
$=0.5\times \left(\frac{5}{60}\right)$
or $f_L=\frac{120}{5}=24cm$
Focal length for mirror,
$f_m=\frac{R}{2}=\frac{-20}{2}=-10cm$
$\therefore$ Equivalent focal length
$f_{eq}=\frac{-2}{f_L}+\frac{1}{f_m}=\frac{-2}{24}+\frac{1}{-10}=\frac{-11}{60}$
$\Rightarrow f_{eq}=\frac{-60}{11}cm$
Hence, this system behaves like a concave mirror of focal length $\frac{-60}{11}cm$