Question 6

**Setting up the effective geometry** The source S sits at depth t/2 inside the slab (index n₀). Seen from either mirror, S appears shifted toward that mirror by Δ = (t/2)(1 − 1/n₀). So the apparent (effective) distance of S from each mirror is d = D/2 − (t/2)(1 − 1/n₀), where D is the mirror separation. **Condition 1: each mirror alone returns the light to S** Light from S reflected by one mirror comes back to S if the apparent source is at the centre of curvature: d = 2f ⟹ D/2 = 2f + (t/2)(1 − 1/n₀) ⟹ D = 4f + (1 − 1/n₀)t. This is option (A). **Condition 2: image formed via both mirrors** If the apparent source lies at the focus of the left mirror (d = f), the reflected beam is parallel to the axis. A parallel beam passes through the slab undeviated, strikes the right mirror and converges to its focus, which (by symmetry) is the apparent position of S – i.e. the actual point S: d = f ⟹ D/2 = f + (t/2)(1 − 1/n₀) ⟹ D = 2f + (1 − 1/n₀)t. This is option (B). (A general symmetric analysis with effective separation D_eff = D − t(1 − 1/n₀) gives exactly d = 2f or d = f, i.e. D_eff = 4f or 2f.) Options (C) and (D) have the shift (n₀ − 1)t, which is incorrect – the apparent shift due to a slab is t(1 − 1/n₀), not (n₀ − 1)t. **Final answer: (A) and (B)**

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