Question

The radius of curvature for a convex lens is 40 cm, for each surface. Its refractive index is $ 1.5 $ . The focal length will be
(A) 40 cm
(B) 20 cm
(C) 80 cm
(D) 30 cm

Answer 1

Hint: This question can be solved by the application of the lens maker’s formula. We need to compute the given values in the lens makers’ formula to get the answer. It must be remembered that the radius of curvature magnitude is the same, but the values are opposite to each other.

Formula used: The formulae used in the solution are given here.
The formulae used in the solution are given here.
 $ \dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) $ where $ f $ is the focal length of the mirror, $ \mu $ is the refractive index of water with respect to the lens, $ {R_1} $ is the radius of curvature of the lens surface closer to the light source, $ {R_2} $ is the radius of curvature of the lens surface farther from the light source.

Complete step by step solution:
It has been given that the radius of curvature for a convex lens is 40 cm, for each surface. Its refractive index is $ 1.5 $ .
According to the lens makers’ formula,
 $ \dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) $ where $ f $ is the focal length of the mirror, $ \mu $ is the refractive index of lens with respect to water, $ {R_1} $ is the radius of curvature of the lens surface closer to the light source, $ {R_2} $ is the radius of curvature of the lens surface farther from the light source.
Let $ {R_1} $ and $ {R_2} $ be two surfaces of a convex lens.
From the question, we can write,
 $ {R_1} = + 40cm $ and $ {R_2} = - 40cm $ .
The refractive index of the lens material is $ 1.5 $ and since, the refractive index is represented by $ \mu $ , where $ \mu = 1.5 $ .
Substituting the given values in the lens makers’ formula, we get,
 $ \dfrac{1}{f} = \left( {1.5 - 1} \right)\left( {\dfrac{1}{{40}} - \dfrac{1}{{\left( { - 40} \right)}}} \right) $
Simplifying, we can get,
 $ \dfrac{1}{f} = 0.5 \times \left( {\dfrac{1}{{40}} + \dfrac{1}{{40}}} \right) = \dfrac{1}{2} \times \dfrac{2}{{40}} = \dfrac{1}{{40}}cm $ .
From this result, we get,
 $ \dfrac{1}{f} = \dfrac{1}{{40}}cm $
 $ \Rightarrow f = 40cm $
Thus, the focal length of the given convex lens is 40cm.
Hence, the correct answer is Option D.

Note:
A convex lens is also known as a converging lens. A converging lens is a lens that converges rays of light that are travelling parallel to its principal axis. They can be identified by their shape which is relatively thick across the middle and thin at the upper and lower edges.
As light approaches the lens, the rays are parallel. As each ray reaches the glass surface, it refracts according to the effective angle of incidence at that point of the lens. Since the surface is curved, different rays of light will refract to different degrees; the outermost rays will refract the most. This runs contrary to what occurs when a divergent lens (otherwise known as concave, biconcave or plano-concave) is employed. In this case, light is refracted away from the axis and outward.