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}} =(\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} $
$\left(\because f_{a}=\frac{1}{p}=-\frac{1}{5} m \right)$
$=1.6 \,m$
$\therefore P_{m} =\frac{\mu}{f_{m}}$
$=\frac{1.6}{1.6}=1\, D$