Water \( \left(\mu_w=\frac{4}{3}\right) \) is filled in a beaker up to a height of 10 cm. A plane mirror is fixed horizontally 5 cm above the water surface, facing down. How far from the mirror is the image of an object O at the bottom of the beaker, seen after reflection in the mirror?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 12.5 cm
The light from O first refracts at the water surface. So the mirror receives it as if it came from the apparent position of O. Viewed near the normal, the apparent depth is the real depth divided by the refractive index of water. The plane mirror then forms the image as far behind it as this apparent object is in front of it.
The apparent depth of O below the water surface is:
\( \begin{aligned} h_1 &= \frac{h_2}{\mu_w} \\ &= \frac{10\,\mathrm{cm}}{4/3} \\ &= 7.5\,\mathrm{cm} \end{aligned} \)
The mirror is 5 cm above the surface, so the apparent object is at a distance from the mirror of:
\( \begin{aligned} d &= 7.5\,\mathrm{cm}+5\,\mathrm{cm} \\ &= 12.5\,\mathrm{cm} \end{aligned} \)
Therefore, the image formed by the mirror is 12.5 cm behind the mirror, that is, 12.5 cm from it.
NCERT (2026-27):
Refraction – “… For viewing near the normal direction, it can be shown that the apparent depth (h1) is real depth (h2) divided by the refractive index of the medium (water).”