Question

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

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
$P^{\prime }=\frac{1}{f^{\prime }}=\left(\frac{{\mu }_g}{{\mu }_l}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\frac{P^{\prime }}{P}=\frac{\left(\frac{{\mu }_g}{{\mu }_l}-1\right)}{\left(\frac{{\mu }_g}{{\mu }_a}-1\right)}=\frac{\left(\frac{1.5}{1.6}-1\right)}{\left(\frac{1.5}{1}-1\right)}$
$=\frac{\left(-0.1/1.6\right)}{0.5}$
$P^{\prime }=-\frac{1\times 2}{16\times 1}P$
$=-\frac{1}{8}\left(-8D\right)$
$=1D$