Question

A plano-convex lens is made of material of refractive index $1.6$. The radius of curvature of the curved surface is $60\,cm.$ The focal length of the lens is
$\left( A \right)\,\,50\,cm \\\ \left( B \right)\,\,200\,cm \\\ \left( C \right)\,\,100\,cm \\\ \left( D \right)\,\,400\,cm \\\$

Answer 1

From the question, the refractive index and radius of curvature of the curved surface are given. By substituting the value of refractive index and the radius in the equation of lens maker formula for convex lens we calculate the value of focal length of the lens.

Formulae Used:
The expression for finding the focal length of the lens is
$\dfrac{1}{f}\, = \,\left( {\mu \, - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Where,
${R_1}$ be the radius of curvature of plane surface, ${R_2}$ be the radius of curvature of curved convex surface, $\mu$ be the refractive index of the lens and $f$ be the focal length of the lens.

Complete step-by-step solution :
We know that,
Radius of curvature of curved convex surface ${R_2} = \,60\,cm$
Refractive index of the lens $\mu = \,{1.6_{}}$
$\dfrac{1}{f}\, = \,\left( {\mu \, - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)..........\left( 1 \right)$
Substitute the known values in the equation $\left( 1 \right)$
$\dfrac{1}{f}\, = \,\left( {1.6\, - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{60}}} \right)$
Let Radius of curvature of plane surface ${R_1} = \infty$
$\dfrac{1}{f}\, = \,\left( {1.6\, - 1} \right)\left( {\dfrac{1}{\infty } - \dfrac{1}{{60}}} \right)$
Simplify the equation, we get
$\dfrac{1}{f}\, = \,\left( {0.6} \right)\left( { - \dfrac{1}{{60}}} \right)$
$\dfrac{1}{f}\, = \, - {\dfrac{1}{{100}}_{_{}}}$
Hence from the above option, option $\left( C \right)$ is correct.
Therefore, focal length of the lens is $f = 100\,cm$.

Note:- We know that the focal length of a convex lens is positive and a focal length of concave lens is negative. Power of a lens is given as (1/f). Thus, for the convex lens, if the focal length is positive then the power is also positive. But for a concave lens, it is negative.