Question

The refractive index of a material of a planoconcave lens is $5/3$, the radius of curvature is $0.3m$. The focal length of the lens in air is

• A. $-0.45m$
• B. $-0.6m$
• C. $-0.75m$
• D. $-1.0m$

Answer 1

The correct option is A $-0.45m$

Lens maker's formula $=\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$
where, $R_2=\infty ,R_1=-0.3m$
$\therefore \frac{1}{f}=(\frac{5}{3}-1)(\frac{1}{-0.3}-\frac{1}{\infty })$
$\Rightarrow \frac{1}{f}=\frac{2}{3}\times \frac{1}{0.3}$
Or $f=0.45m$