Question

A thin equiconvex lens of refractive index $\frac{3}{2}$ and radius of curvature $30cm$ is put in water (refractive index = $\frac{4}{3})$, its focal length is

• A. $0.15m$
• B. $0.30m$
• C. $0.45m$
• D. $1.20m$

Answer 1

Answer: (D)

According to lens maker's formula
$\frac{1}{f}=\left(\mu -1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
where $\mu =\frac{{\mu }_L}{{\mu }_M}$
Here, ${\mu }_L=\frac{3}{2}$, ${\mu }_M=\frac{4}{3}$, $R_1=+30cm$, $R_2=-30cm$
$\therefore \frac{1}{f}=\left(\frac{\frac{3}{2}}{\frac{4}{3}}-1\right)\left(\frac{1}{30}-\frac{1}{-30}\right)$
$=\left(\frac{1}{8}\right)\left(\frac{2}{30}\right)$
$\frac{1}{f}=\frac{1}{4\times 30}=\frac{1}{120}$
or $f=120cm=1.2m$