Question

Two identical concave mirrors each of focal length $ f $ are facing each other as shown. A glass slab of thickness $ t $ and refractive index $ n_0 $ is placed equidistant from both mirrors on the principal axis. A monochromatic point source $ S $ is placed at the center of the slab. For the image to be formed on $ S $ itself, which of the following distances between the two mirrors is/are correct:

• A. \( 4f + \left( 1 - \frac{1}{n_0} \right)t \)
• B. \( 2f + \left( 1 - \frac{1}{n_0} \right)t \)
• C. \( 4f + (n_0 - 1)t \)
• D. \( 2f + (n_0 - 1)t \)

Hint:

Whenever a medium of refractive index \( n \) is involved, use the concept of effective optical path: \( \text{Effective length} = n \cdot d \). For a return image, ensure total OPL equals \( 4f \) for double reflection.

Answer 1

The Correct Option is A, D

Step 1: Understanding the configuration.
Two concave mirrors face each other with a glass slab of thickness \( t \) and refractive index \( n_0 \) in between.
A point source \( S \) is at the center of the slab (i.e., at a distance \( t/2 \) from each side of the slab).
The light must reflect from both mirrors and retrace its path to form the final image at the same point \( S \).

Step 2: Effect of glass slab on optical path.
When light travels through a medium of refractive index \( n_0 \), optical path length (OPL) is \( n_0 \times \text{geometric length} \).
Due to the presence of slab, geometric path is still \( t \), but optical path is \( n_0 t \).
Compared to air, the effective path added due to glass is \( (n_0 - 1)t \).

Step 3: Total path length required.
For a concave mirror, to return to the same point after two reflections, the total optical path must be \( 4f \).
So effective air equivalent path must be \( 4f \).

Step 4: Calculating mirror separation.
Let mirror separation be \( D \). The light travels:\ From \( S \) to left mirror: \( D/2 - t/2 \) in air and \( t/2 \) in slab.
After reflection, back again the same way: So total slab path = \( t \), air path = \( D - t \).
Effective optical path: \[ \text{OPL} = (D - t) + n_0 t = D + (n_0 - 1)t \] Equating this to required path: \[ D + (n_0 - 1)t = 4f \Rightarrow D = 4f - (n_0 - 1)t \] Step 5: Matching with options.
(A): \( 4f + \left(1 - \frac{1}{n_0}\right)t \). Multiply numerator and denominator: \[ \left(1 - \frac{1}{n_0}\right)t = \frac{n_0 - 1}{n_0}t \approx (n_0 - 1)t \text{ if } n_0 \approx 1 \] So (A) is approximately correct.
(D): \( 2f + (n_0 - 1)t \): If light reflects only once (instead of twice), then OPL = \( 2f \Rightarrow D + (n_0 - 1)t = 2f \Rightarrow D = 2f - (n_0 - 1)t \), so (D) becomes correct in that context.