Question

The radii of curvature of the two surfaces of a lens are $20cm$ and $30cm$ and the refractive index of the material of the lens is $\mathrm{1.5.}$ If the lens is concavo-convex, then the focal length of the lens is

• A. 24 cm
• B. 10 cm
• C. 15 cm
• D. 120 cm

Answer 1

Answer: (D)

The focal length of the lens
$\frac{1}{f}=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\frac{1}{f}=(1.5-1)\left(\frac{1}{20}-\frac{1}{30}\right)$
$\frac{1}{f}=0.5\left(\frac{30-20}{600}\right)$
or $\frac{1}{f}=\frac{1}{2}\times \frac{10}{600}=\frac{1}{120}$
or $f=120cm$