Question

If radii of curvature of both convex surfaces is $20cm$, then focal length of the lens for an object placed in air in the given arrangement is

• A. $10cm$
• B. $20cm$
• C. $40cm$
• D. $80cm$

Answer 1

Answer: (C)

The correct option is C $40cm$
Given: If radii of curvature of both convex surfaces is 20 cm,

To find the focal length of the lens for an object placed in air in the given arrangement

Solution:

Using the formula for the refraction through curved surface for first surface, we get

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}⟹\frac{1.5}{v^{\prime }}-\frac{1}{u}=\frac{1.5-1}{20}.....\left(i\right)$

Using the formula for the refraction through curved surface for second surface, we get

$\frac{{\mu }_3}{v}-\frac{{\mu }_2}{v^{\prime }}=\frac{{\mu }_2-{\mu }_1}{R}⟹\frac{\frac{4}{3}}{v}-\frac{1.5}{v^{\prime }}=\frac{\frac{4}{3}-1.5}{-20}.....\left(ii\right)$

Adding (i) and (ii),we get

$\frac{1.5}{v^{\prime }}-\frac{1}{u}+\frac{\frac{4}{3}}{v}-\frac{1.5}{v^{\prime }}=\frac{1.5-1}{20}+\frac{\frac{4}{3}-1.5}{-20}⟹-\frac{1}{u}+\frac{\frac{4}{3}}{v}=\frac{0.5}{20}+\frac{(-0.167)}{-20}$

For, $u=\infty$ and $v=f$

$⟹-\frac{1}{\infty }+\frac{\frac{4}{3}}{f}=\frac{0.5+0.167}{20}⟹\frac{4}{3f}=\frac{0.667}{20}⟹f=\frac{4\times 20}{3\times 0.667}⟹f=40cm$

is the focal length of the lens for an object placed in air in the given arrangement