Question

The refractive index of a lens material is $ \mu $ and focal length $ f $ . Due to some chemical changes in the material, its refractive index has increased by $ 2\% $ . The percentage decrease, in focal length for $ {{\mu = 1}}{{.5}} $ will be:
(A) $ 4\% $
(B) $ 2\% $
(C) $ 5.6\% $
(D) $ 8\% $

Answer 1

Hint: This question can be solved by the application of the lens maker’s formula. The decrease in focal length can be calculated by finding the ratio between the new refractive index and the original refractive index.

Formula Used: The formulae used in the solution are given here.
 $ \dfrac{1}{f} = \left( {\eta - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) $ where f is the focal length of the mirror, $ \eta $ 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
Lens maker’s formula is the relation between the focal length of a lens to the refractive index of its material and the radii of curvature of its two surfaces. It is used by lens manufacturers to make the lenses of particular power from the glass of a given refractive index.
According to the lens makers’ formula,
 $ \dfrac{1}{f} = \left( {\eta - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) $ where f is the focal length of the mirror, $ \eta $ 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.
From this formula, we come to know that, $ f\alpha \dfrac{1}{{\eta - 1}} $ .
It has been given that the refractive index of a lens material is $ \mu $ and focal length $ f $ . Due to some chemical changes in the material, its refractive index has increased by $ 2\% $ .
 $ \dfrac{1}{f} = \left[ {\mu - 1} \right]\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) $ .
Let $ f' $ be the new focal length for refractive index $ \mu ' $ . Thus,
 $ \dfrac{1}{{f'}} = \left[ {\mu ' - 1} \right]\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) $ .
It has been given that, $ \mu ' = \mu + \dfrac{2}{{100}}\mu $ .
It has been stated that, $ {{\mu = 1}}{{.5}} $ . Substituting the value of $ \mu $ , we get,
 $ \mu = 1.5 + \dfrac{2}{{100}} \times 1.5 $
 $ \Rightarrow \mu = 1.53 $
Taking the ratio of $ f' $ and $ f $ , we get,
 $ \dfrac{f}{{f'}} = \dfrac{{\mu - 1}}{{\mu ' - 1}} $ where $ f' $ and $ \mu ' $ are the new focal length and refractive index.
 $ \dfrac{f}{{f'}} = \dfrac{{1.53 - 1}}{{1.5 - 1}} = \dfrac{{0.53}}{{0.5}} $ .
Simplifying the equation above,
 $ \dfrac{{f'}}{f} = \dfrac{{50}}{{53}} $
 $ \Rightarrow f' = \dfrac{{50}}{{53}}f $
As we can understand from the equation above, the focal length clearly decreases.
The change in percentage in focal length for $ {{\mu = 1}}{{.5}} $ will be given by,
 $ {{\Delta f in \% }} = \dfrac{{f - \dfrac{{50}}{{53}}f}}{f} \times 100 \simeq 5.64\% $ where $ {{\Delta f }} $ is the change in focal length.
Hence, the correct answer is Option D.

Note
Real lenses have the finite thickness between their two surfaces of curvature. An ideal thin lens with two surfaces of equal curvature will have zero optical power. It means it will neither converge or diverge light. A lens with some thickness which is not negligible is called a thick lens. Lenses are of two types based on the curvature of the two optical surfaces. Convex and concave.
Lens maker’s formula is the relation between the focal length of a lens to the refractive index of its material and the radii of curvature of its two surfaces. It is used by lens manufacturers to make the lenses of particular power from the glass of a given refractive index.
Lens maker formula is used to construct a lens with the specified focal length. A lens has two curved surfaces, but these are not exactly the same. If we know the refractive index and the radius of the curvature of both the surface, then we can determine the focal length of the lens.