Question

The image of an object kept at a distance of 20 cm in front of a concave mirror is found to coincide with itself. If a glass slab ($\mu$=1.5) of thickness 3 cm is introduced between the mirror and the object, then in order that the final image again coincides with the object:

• A. the mirror should be displaced away from the object
• B. the mirror should be displaced towards the object
• C. the magnitude of displacement is 1 cm
• D. the magnitude of displacement is 0.5 cm

Answer 1

Answer: (A), (C)

the mirror should be displaced away from the object and the magnitude of displacement is 1 cm

Answer 2

Initially, the object is at the center of curvature (C) of the concave mirror, so R = 20 cm.

When a glass slab of thickness t = 3 cm and refractive index $\mu$ = 1.5 is introduced, the object appears to shift towards the mirror by a distance s, given by:

$s = t(1 - \frac{1}{\mu}) = 3(1 - \frac{1}{1.5}) = 3(1 - \frac{2}{3}) = 3(\frac{1}{3}) = 1$ cm.

So, the object appears to be at a distance of 20 - 1 = 19 cm from the mirror.

To make the image coincide with the object again, the object must be at the center of curvature (20 cm). Let the mirror be displaced by $\Delta x$.

If the mirror is displaced away from the object:

Effective distance = $(20 + \Delta x) - 1 = 19 + \Delta x$.

We need $19 + \Delta x = 20$, so $\Delta x = 1$ cm.

If the mirror is displaced towards the object:

Effective distance = $(20 - \Delta x) - 1 = 19 - \Delta x$.

We need $19 - \Delta x = 20$, so $\Delta x = -1$ cm (which means away from the object).

Therefore, the mirror should be displaced away from the object by 1 cm.