Question

There is an equiconvex lens of focal length of 40 cm. The lens is cut into two equal parts perpendicular to the principal axis, the focal lengths of each part will be:
(A). 30 cm
(B). 10 cm
(C). 40 cm
(D). 80 cm

Answer 1

- Hint: When we will cut the lens in two equal parts , the radius of splitted parts will be the same but the sign will be different. The focal length of the individual pieces can be obtained from the lens maker formula.

Complete step-by-step solution -
The lens maker formula of a thin lens given by,
$\dfrac{1}{f}=(n-1)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$
Here, $f$ is the focal length, ${{R}_{1}},\,{{R}_{2}}$ are the radii of curvature of the two curved surfaces and is the refractive index of the material. For an equiconvex lens, the radii are equal. Taking into account the sign convention,
${{R}_{1}}=-{{R}_{2}}=R$
$R$ is the radius of curvature of each surface. Hence, from the lens maker formula,
$\dfrac{1}{f}=(n-1)\left( \dfrac{1}{R}-\dfrac{1}{-R} \right)$.
$\dfrac{1}{f}=(n-1)\dfrac{2}{R}$.
$\dfrac{1}{2f}=(n-1)\dfrac{1}{R}$
Here, $f=40\,\,\text{cm}$.
The individual lenses are plano-convex lenses with radii of curvature being $R$ and $\infty$. From the lens maker equation,
$\dfrac{1}{{{f}_{in}}}=(n-1)\left( \dfrac{1}{R}-\dfrac{1}{\infty } \right)$
$\dfrac{1}{{{f}_{in}}}=(n-1)\dfrac{1}{R}$
Here, ${{f}_{in}}$ is the focal length of the individual parts. Using equation (1),
\[\dfrac{1}{{{f}_{in}}}=(n-1)\dfrac{1}{R}=\dfrac{1}{2f}\]
\[{{f}_{in}}=2f\]
\[{{f}_{in}}=2\times 40\,\,\text{cm}\]
\[{{f}_{in}}=80\,\,\text{cm}\]
Hence, the focal length of each part is 80 cm (option D).

Note: According to the sign convention of thin lenses, distances are measured from the center of a lens. Distances in the direction of the light and above the principal axis are taken to be positive.