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

Answer 1

Answer: (A)

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