Question

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

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

Answer 1

Answer: (A)

Lens maker's formula
$\frac{1}{f}=\left(\mu - 1\right) \, \left[\frac{1}{R_{1}} - \frac{1}{R_{2}}\right]$
Where , $R_{2}= \, \in fty, \, R_{1}=0.3 \, m$
$\therefore \frac{1}{f}=\left(\frac{5}{3} - 1\right)\left(\frac{1}{0.3} - \frac{1}{\in fty}\right)$
$\Rightarrow \frac{1}{f}=\frac{2}{3}\times \frac{1}{0.3}$
$Or \, f=0.45 \, m$