A glass prism of angle \( 60^\circ \) has an angle of minimum deviation of \( 30^\circ \) for a given colour of light. The refractive index of glass for that colour of light is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.414
At minimum deviation the ray passes through the prism symmetrically, and the refractive index is given by the prism formula
\( \displaystyle n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \)
\( A = 60^\circ \)
\( \delta_m = 30^\circ \)
\( \displaystyle n = \frac{\sin\left(\frac{60^\circ + 30^\circ}{2}\right)}{\sin 30^\circ} = \frac{\sin 45^\circ}{\sin 30^\circ} \)
\( \displaystyle n = \frac{1/\sqrt{2}}{1/2} = \sqrt{2} = 1.414 \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
