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 $1.5$. If the lens is concavo-convex then the focal length of the lens is:

• A. $24cm$
• B. $10cm$
• C. $120cm$
• D. $-120cm$

Answer 1

The correct option is D $-120cm$

From lens maker's formula, $\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

We have,

Refractive index, $\mu =1.5$ and $R_1=20cm$ and $R_2=30cm$

$\frac{1}{f}=(1.5-1)(\frac{1}{R_1}-\frac{1}{R_2})$

$\frac{1}{f}=(1.5-1)(\frac{1}{-20}-\frac{1}{-30})$

$\frac{1}{f}=(1.5-1)\frac{-30+20}{20\times 30}$

$f=-120cm$