Question

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

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

Answer 1

Answer: (D)

$1.20 \,m$

Answer 2

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, $\mu_{L}=\frac{3}{2}$, $\mu_{M}=\frac{4}{3}$, $R_{1}=+30\,cm$, $R_{2}=-30\,cm$ $\therefore \frac{1}{f}=\left(\frac{\frac{3}{2}}{\frac{4}{3}}-1\right)\left(\frac{1}{30}-\frac{1}{-30}\right)$ $=\left(\frac{1}{8}\right)\left(\frac{2}{30}\right)$ $\frac{1}{f}=\frac{1}{4\times30}=\frac{1}{120}$ or $f=120\,cm=1.2\,m$