Question

A thin glass (refractive index $1.5$) lens has optical power of $-5D$ in air. Its optical power in a liquid medium with refractive index $1.6$ will be

• A. 1 D
• B. -1 D
• C. 25 D
• D. -25 D

Answer 1

Answer: (A)

$\frac{1}{f_a}=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$=(1.5-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ ... (i)
and $\frac{1}{f_m}=\left(\frac{{\mu }_g-{\mu }_m}{{\mu }_m}\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\frac{1}{f_m}=\left(\frac{1.5}{1.6}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ ... (ii)
Thus, $\frac{f_m}{f_a}=\frac{(1.5-1)}{\left(\frac{1.5}{1.6}-1\right)}=-8$
$f_m=-8\times f_a$
$=-8\times \frac{-1}{5}$
$\left(\because f_a=\frac{1}{p}=-\frac{1}{5}m\right)$
$=1.6m$
$\therefore P_m=\frac{\mu }{f_m}$
$=\frac{1.6}{1.6}=1D$