Question

An equiconvex lens has a power of $5$ dioptre. If it is made of glass of refractive index $1.5$, then the radius of curvature of its each surface will be?
A) $20 cm$
B) $10 cm$
C) $40 cm$
D) $\infty$

Answer 1

Hint: Power of a lens is given and refractive index of glass is also given. We can find the focal length by the use of power as it is equal to the inverse of the power. In an equiconvex lens, the radius of curvature of both the surfaces are equal in magnitude. Using the lens maker formula, we can determine the radius of curvature of each surface.

Complete step-by-step solution:
Given: Power of equiconvex lens, P = $5$ dioptre.
Focal length, f, is the reciprocal of power.
$P = \dfrac{1}{f}$
Refractive index of glass, n = $1.5$
Using Lens Maker Formula,
$\dfrac{1}{f} = (n-1) \times \left( \dfrac{1}{R_{1}} - \dfrac{1}{R_{2}} \right)$
Where, f is the focal length of the lens.
n is the refractive index.
$R_{1}$ and $R_{2}$ are the radius of curvature of both the surfaces.
For an equiconvex lens, $R_{1} =R$ and $R_{2}=-R$. So, lens maker formula will be,
$\dfrac{1}{f} = (n-1) \times \left( \dfrac{1}{R} + \dfrac{1}{R} \right)$
$\implies \dfrac{1}{f} = (n-1) \times \left( \dfrac{2}{R} \right)$
Put the values of n and f.
$5 = (1.5-1) \times \left( \dfrac{2}{R} \right)$
$\implies R = 0.2 m$
$\implies R = 20 cm$
Option (a) is right.

Note: The lens whose both refracting covers have an equal radius of curvature is named an equiconvex lens. A convex lens is thick at the middle and thin at the edges. It is a converging lens which means that it converges light rays after refraction. A concave lens is thick at the edges and thin at the edges, known as a concave lens. It is diverging in nature, which implies that it diverges light rays after refraction.