Question

The equiconvex lens has focal length, $f$ . If it is cut perpendicular to the principal axis passing through optical centre, then focal length of each half is:
A. $\dfrac{f}{2}$
B. $f$
C. $\dfrac{{3f}}{2}$
D. $2f$

Answer 1

In order to this question, to find the focal length of each half after the convex lens is cut perpendicular to the principal axis passing through optical centre, we will first apply the Lens Maker formula and then again we will apply equi-convex lens and then we will apply plano-convex lens, as it is cut perpendicularly.

Complete step by step answer:
Let the two spheres be sphere 1 and sphere 2 whose radius of curvature are ${R_1}$ and ${R_2}$ respectively. Here, we will go through the Lens Maker’s Formula:-
$\dfrac{1}{f} = (\mu - 1)(\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}})$
here, $f$ is the focal length (half the radius of curvature), $\mu$ is the refractive index of the material used, ${R_1}$ is the radius of curvature of sphere 1 and ${R_2}$ is the radius of curvature of sphere 2.

Now, for equi-convex lens:- (in which radius of curvature of both the spheres are same)
$\because \dfrac{1}{f} = (\mu - 1)(\dfrac{1}{R} - \dfrac{1}{{ - R}}) \\\ \Rightarrow \dfrac{1}{f} = (\mu - 1)\dfrac{2}{R} \\\$
So, according to the question, if the convex lens is cut perpendicular to the principal axis passing through the optical centre, there is the formation of plano-convex lens. And we will apply the formula of plano-convex lenses. Again, for the plano-convex lens formed:-
$\dfrac{1}{{{f^`}}} = (\mu - 1)(\dfrac{1}{{{R_1}}} - \dfrac{1}{\infty }) \\\ \Rightarrow \dfrac{1}{{{f^`}}} = (\mu - 1)\dfrac{1}{R} \\\ \therefore {f^`} = 2f \\\$
Hence, the correct option is D.

Note: If we only have the refractive index $n$ , thickness $t$ , and aperture diameter $d$ , we won't be able to calculate the radius. The reason for this is that two lenses with equal $n,t\,and{\text{ }}d$ values may exist, but their edge thicknesses would differ.