Question 6

Let the distance between the two mirrors be $L$. The source $S$ is at the center, so its physical distance from each mirror is $L/2$. Because the source is inside a slab of thickness $t$, there is a glass medium of thickness $t/2$ between the source and each mirror. The rest of the space $(\frac{L-t}{2})$ is air. The apparent distance $u$ of the source from the mirror is given by:

$u = \frac{L-t}{2} + \frac{t/2}{n_0} = \frac{L}{2} - \frac{t}{2}(1 - \frac{1}{n_0})$.

For the image to be formed on the source itself, two conditions are possible:

1). The rays strike the mirror normally and return along the same path. This happens if the apparent position of $S$ is at the center of curvature ($u = R = 2f$).

Thus, $\frac{L}{2} - \frac{t}{2}(1 - \frac{1}{n_0}) = 2f \Rightarrow L = 4f + (1 - \frac{1}{n_0})t$. This matches option (A).

2). The rays from $S$ become parallel after reflection from the first mirror ($u = f$). These parallel rays then strike the second mirror and, due to symmetry, focus at the same point $S$ (since $S$ is also at the apparent focus of the second mirror).

Thus, $\frac{L}{2} - \frac{t}{2}(1 - \frac{1}{n_0}) = f \Rightarrow L = 2f + (1 - \frac{1}{n_0})t$. This matches option (B).