Question

A thin glass (refractive index $1.5$) lens has optical power of $-8\, D$ 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)

In air,
$P =\frac{1}{f}=\left(\frac{\mu_{g}}{\mu_{a}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) $
In medium
$ P' =\frac{1}{f'} = \left(\frac{\mu_{g}}{\mu_{l}} -1\right)\left(\frac{1}{R_{1} }-\frac{1}{R_{2}}\right) $
$\frac{P'}{P} = \frac{\left(\frac{\mu_{g}}{\mu_{l}}-1\right)}{\left(\frac{\mu_{g}}{\mu_{a}}-1\right)} = \frac{\left(\frac{1.5}{1.6}-1\right)}{\left(\frac{1.5}{1}-1\right)} $
$= \frac{\left(-0.1/1.6\right)}{0.5} $
$P' = -\frac{1\times 2}{16\times 1} P $
$= -\frac{1}{8}\left(-8 \,D\right)$
$ = 1\,D$