Two identical concave mirrors each of focal length are facing each other as shown in the schematic diagram. The focal length is much larger than the size of the mirrors. A glass slab of thickness and refractive index is kept equidistant from the mirro

Question

Two identical concave mirrors each of focal length $f$ are facing each other as shown in the schematic diagram. The focal length $f$ is much larger than the size of the mirrors. A glass slab of thickness $t$ and refractive index $n_{0}$ is kept equidistant from the mirrors and perpendicular to their common principal axis. A monochromatic point light source $S$ is embedded at the center of the slab on the principal axis, as shown in the schematic diagram. For the image to be formed on $S$ itself, which of the following distances between the two mirrors is/are correct [2025]

Smart Achievers · Accepted Answer

(1, 2)

From the figure, $f$

$S + t (\frac{1}{n_{0}} - 1) = 4 f$

$S = 4 f + (1 - \frac{1}{n_{0}}) t$

So, option (1) is correct.

Also, $\frac{S - t}{2} + \frac{t}{2 n_{0}} = f$

$S = 2 f + (1 - \frac{1}{n_{0}}) t$

Hence, option (2) is correct.

Solution

1 $4 f + (1 - \frac{1}{n_{0}}) t$

2 $2 f + (1 - \frac{1}{n_{0}}) t$

3 $4 f + (n_{0} - 1) t$

4 $2 f + (n_{0} - 1) t$

Ans.
(1, 2)

From the figure, $\frac{S - t}{2} + \frac{t}{2 n_{0}} = 2 f$

$S + t (\frac{1}{n_{0}} - 1) = 4 f$

$S = 4 f + (1 - \frac{1}{n_{0}}) t$

So, option (1) is correct.

Also, $\frac{S - t}{2} + \frac{t}{2 n_{0}} = f$

$S = 2 f + (1 - \frac{1}{n_{0}}) t$

Hence, option (2) is correct.

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Solution

1 4f+(1-1n0)t

2 2f+(1-1n0)t

3 4f+(n0-1)t

4 2f+(n0-1)t