Question

The figure shows a thin lens with point $C_1$ and $C_2$ as the centres of curvature.
The focal length of lens will be (refractive index = 1.5)

• A. 60 cm
• B. 45 cm
• C. 120 cm
• D. 75 cm

Answer 1

Answer: (A)

The radii of the curvature are positive Thus, $R_1=+15cm$ and $R_2=+20cm$ The focal length is given by $\frac{1}{f}=(\mu -1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$ $\Rightarrow$ $\frac{1}{f}=(1.5-1)\left[\frac{1}{15}-\frac{1}{20}\right]$ $=0.5\times \frac{1}{60}=\frac{5}{600}$ $\Rightarrow$ $f=\frac{600}{5}=120cm$