Question

: A biconcave lens has a focal length of $\dfrac{2}{3}$ times the radius of curvature of either surface. The refractive index of the lens material is
(A) $1.75$
(B) $1.33$
(C) $1.5$
(D) $1.0$

Answer 1

Since the lens is bi-convex, then the radius of curvature is equal on the two sides. The lens maker equation should be used by us in solving this problem (as this is not necessarily a thin lens).

Formula used: In this solution we will be using the following formula;
$\dfrac{1}{f} = \left( {n - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ where $f$ is the focal length, $n$ is the refractive index of the lens, ${R_1}$ is the radius of curvature on one side and ${R_2}$ is the radius of curvature on the other side.

Complete step by step answer
The focal length of a biconvex lens is said to be $\dfrac{2}{3}$ of the radius of curvature. Hence, to solve this, the lens maker equation must be used. This is given by
$\dfrac{1}{f} = \left( {n - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ where $f$ is the focal length, $n$ is the refractive index of the lens, ${R_1}$ is the radius of curvature on one side and ${R_2}$ is the radius of curvature on the other side.
Now, since the lens is biconvex, it has the same radius on the two sides, but one of them is usually considered negative, say ${R_2}$ i.e. ${R_1} = - {R_2} = R$
Hence, substituting known values we have
$\dfrac{1}{{\left( {\dfrac{2}{3}R} \right)}} = \left( {n - 1} \right)\left( {\dfrac{1}{R} - \dfrac{1}{{ - R}}} \right)$
By simplifying,
$\dfrac{3}{{2R}} = \left( {n - 1} \right)\left( {\dfrac{1}{R} + \dfrac{1}{R}} \right)$
$\Rightarrow \dfrac{3}{{2R}} = \left( {n - 1} \right)\dfrac{2}{R}$
Cancelling $R$ from both sides and making $n$ subject of formula, we have
$\dfrac{3}{4} = n - 1$
$\Rightarrow n = \dfrac{3}{4} + 1 = \dfrac{7}{4}$
Hence, the refractive index is $n = 1.75$

Thus, the correct option is A.

Note
For clarity, we could tell that the lens is not thin because for a thin lens, the focal length is always equal to about half of the radius of curvature, i.e. $f = \dfrac{1}{2}R$.
Also, actually, the lens maker equation is written as
$\dfrac{1}{f} = \left( {\dfrac{{{n_l}}}{{{n_i}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ Where ${n_l}$ is the refractive index of the lens, and ${n_i}$ is the refractive index of the medium the lens is placed. Hence in the above, it is assumed air and ${n_i} = 1$