Question

A thin glass (refractive index 1.5) lens has optical power of $- 5\, D$ 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)

$\frac{1}{f_{a}}=\left(\mu-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$
$=\left(1.5-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\,...\left(i\right)$
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, $\frac{f _{m}}{f_{a}}=\frac{\left(1.5-1\right)}{\left(\frac{1.5}{1.6}-1\right)}=-8$
$f _{m}=-8\times f_{a}
=-8\times\frac{-1}{5}\,\left(\because f_{a}=\frac{1}{p}=-\frac{1}{5}\,m\right)$
$=1.6\,m$
$\therefore P_{m}=\frac{\mu}{f _{m}}=\frac{1.6}{1.6}=1\,D$