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, $\mu = 1.5, R_1 = 20 \,cm, R_2 = -20\,cm$
$\therefore \frac{1}{f_{L}} = \left(1.5 -1\right)\left(\frac{1}{20}-\frac{1}{-30}\right)$
$= 0.5 \times\left(\frac{5}{60}\right)$
or $f_{L}= \frac{120}{5} = 24\, cm $
Focal length for mirror,
$f_{m} = \frac{R}{2} = \frac{-20}{2} = -10\, cm$
$ \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 $\frac{-60}{11}\, cm$