1. Analyzing the Optical Path

The light travels from the object, passes through the lens, reflects off the plane mirror, and passes through the lens again. Let the lens be at $x=0$ and the mirror at $x=d$.

• Pass 1 (Lens): Object at $u$. Image formed at $v_1$. Magnification $m_1 = \frac{f}{f+u}$.
• Reflection (Mirror): The image $I_1$ acts as a virtual object for the mirror. The mirror forms an image $I_2$ such that it is symmetric. The magnification of a plane mirror is $m_2 = -1$ (lateral inversion/direction change).
• Pass 2 (Lens): The reflected light travels back through the lens.

2. Calculating Total Magnification

Let’s trace the magnification through the system. The position of the first image $v_1$ is given by: $$ v_1 = \frac{uf}{u+f} $$ This image $I_1$ is at distance $(d – v_1)$ from the mirror. The mirror forms image $I_2$ at distance $(d – v_1)$ behind the mirror. So, the distance of $I_2$ from the lens (for the return trip) is: $$ u’ = d + (d – v_1) = 2d – v_1 $$ However, for the lens in the second pass, the light is incident from the right. We treat $I_2$ as the object. The magnification for the second pass through the lens $m_3$ is: $$ m_3 = \frac{f}{f – (distance\_from\_lens)} = \frac{f}{f – (2d – v_1)} $$ (Note: Sign conventions can be tricky here. A robust method is to require the coefficient of $u$ in the total magnification expression to vanish).

3. Condition for Independence

The total magnification is $M = m_1 \times m_2 \times m_3$. $$ M \propto \left(\frac{f}{f+u}\right) \times \left(\frac{f}{f – 2d + v_1}\right) $$ Substituting $v_1 = \frac{uf}{u+f}$: $$ M \propto \frac{f^2}{(f+u)(f – 2d) + (f+u)v_1} $$ $$ \text{Denominator } D = (f+u)(f-2d) + uf $$ $$ D = f(f-2d) + u(f-2d) + uf $$ $$ D = f(f-2d) + u(f – 2d + f) = f(f-2d) + u(2f – 2d) $$ For the magnification to be independent of $u$, the term containing $u$ in the denominator must be zero. $$ 2f – 2d = 0 \implies d = f $$