A pin on a table is seen from above, near the normal. A glass slab 12 cm thick (refractive index 1.5) is placed flat on the pin. The slab is then lifted 5 cm above the table and held parallel to it. Now the pin appears to be raised by
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4 cm
Near the normal, a thickness \( t \) of glass looks only \( t/n \) thick. The air gap below the slab looks the same with or without the slab. So the pin appears raised by \( t-t/n \), wherever the slab is held.
\( t=12\ \mathrm{cm} \)
\( n=1.5 \)
\( \begin{aligned} \text{shift} &= t-\frac{t}{n} \\ &= 12-\frac{12}{1.5} \\ &= 4\ \mathrm{cm} \end{aligned} \)
The 5 cm air gap does not enter the result. Hence, the pin appears raised by 4 cm.
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).”