Question

When the object is at distance ${u_1}$ and ${u_2}$ from a lens, a real image and virtual image are formed respectively having the same magnification. The focal length of the lens is:
A. ${u_1} - {u_2}$
B. $\dfrac{{{u_1} - {u_2}}}{2}$
C. $\dfrac{{{u_1} + {u_2}}}{2}$
D. ${u_1} + {u_2}$

Answer 1

Hint:In the solution, equate magnification of lens for both the images by using the lens formula and magnification formula for the determination of focal length of the lens.

Complete Step by Step Answer:
The expression of the focal length of the lens for the real image is,
$\dfrac{1}{f} = \dfrac{1}{{{v_1}}} + \dfrac{1}{{{u_1}}}$ (1)
Here, $f$ is the focal length of the lens, ${u_1}$ is the distance of the first object and is the image distance of the first object
The expression of the focal length of the lens for the virtual image is,
$\dfrac{1}{f} = \dfrac{1}{{ - {v_2}}} + \dfrac{1}{{{u_2}}}$ (2)
Here, ${u_2}$ is the distance of the second object, and ${v_2}$ is the image distance of the second object.
Rearrange the equation (1)
$\begin{array}{l}
\dfrac{1}{f} = \dfrac{1}{{{v_1}}} + \dfrac{1}{{{u_1}}}\\
\dfrac{1}{{{v_1}}} = \dfrac{1}{f} - \dfrac{1}{{{u_1}}}\\
{v_1} = \dfrac{{{u_1}f}}{{{u_1} - f}}\\
\dfrac{{{v_1}}}{{{u_1}}} = \dfrac{f}{{{u_1} - f}}
\end{array}$
The ratio of image distance to the object distance is known as magnification, so the above equation becomes,
$m = \dfrac{f}{{{u_1} - f}}$ (3)
Rearrange the equation (2)
$\begin{array}{l}
\dfrac{1}{f} = \dfrac{1}{{ - {v_2}}} + \dfrac{1}{{{u_2}}}\\
\dfrac{1}{{{v_2}}} = \dfrac{1}{{{u_2}}} - \dfrac{1}{f}\\
{v_2} = \dfrac{{{u_2}f}}{{f - {u_2}}}\\
\dfrac{{{v_2}}}{{{u_2}}} = \dfrac{f}{{f - {u_2}}}
\end{array}$
The ratio of image distance to the object distance is known as magnification, so the above equation becomes,
$m' = \dfrac{f}{{f - {u_2}}}$ (4)
The magnification of the real and virtual image is same so from equation (3) and (4), the focal length of the lens is,
$\begin{array}{l}
m = m'\\
\dfrac{f}{{{u_1} - f}} = \dfrac{f}{{f - {u_2}}}\\
{u_1} - f = f - {u_2}\\
f = \dfrac{{{u_1} + {u_2}}}{2}
\end{array}$
Therefore, the option (c) is the correct answer that is $\dfrac{{{u_1} + {u_2}}}{2}$.

Note: Use the correct sign convention with the value of real and virtual images distance in the lens formula for the proper calculations. Use the positive sign for the real image and negative sign for the virtual image.