A microscope is focused on a mark at the bottom of an empty beaker. Water of refractive index 4/3 is then poured into the beaker to a depth of 8 cm. To focus on the mark again, the microscope must be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → raised by 2 cm
An object under water, seen from above near the normal, appears at a smaller depth. This apparent depth is the real depth divided by the refractive index of water. So the mark looks raised, and the microscope has to move up by the same amount to focus on its image again.
The apparent depth of the mark below the water surface is:
\( \begin{aligned} h_1 &= \frac{h_2}{n} \\ &= \frac{8\,\mathrm{cm}}{4/3} \\ &= 6\,\mathrm{cm} \end{aligned} \)
The image of the mark rises by:
\( \begin{aligned} h_2-h_1 &= 8\,\mathrm{cm}-6\,\mathrm{cm} \\ &= 2\,\mathrm{cm} \end{aligned} \)
The image of the mark moves up by this amount. Hence, the microscope must be raised by 2 cm.
NCERT (2026-27):
Refraction – “… Another familiar observation is that the bottom of a tank filled with water appears to be raised …. 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).”