Question

It is given that ${}_a{\mu }_g=$$\frac{3}{2}$ and ${}_a{\mu }_w=$$\frac{4}{3}$. An equi-convex lens with radius of each surface equal to $20cm$ is placed in the water. Then its focal length is

• A. $20cm$
• B. $30cm$
• C. $40cm$
• D. $80cm$

Answer 1

The correct option is D $80cm$

$\frac{1}{f}=(\frac{{\mu }_g}{{\mu }_w}-1)(\frac{1}{R}-\left(\frac{1}{-R}\right))$

$=({}^3{/}_2\times {}^3{/}_4-1)\left(\frac{2}{R}\right)$

=$(\frac{9}{8}-1)\left(\frac{2}{20}\right)$

=$\frac{1}{8}\times \frac{2}{20}$

$f=80cm$