Two beams of red and violet colour are made to pass separately throug…

The correct answer is Option (1) → \( 30^\circ \) for both the colours

For any ray through a prism the two angles of refraction add up to the angle of the prism, \( r_1+r_2=A \). In the position of minimum deviation the ray passes symmetrically, \( i=e \), so \( r_1=r_2 \). Calling this common angle \( r \), \( 2r=A \).

    \( \begin{aligned} r &= \frac{A}{2} \\ &= \frac{60^\circ}{2} \\ &= 30^\circ \end{aligned} \)

Option (1) \( 30^\circ \) for both the colours — Correct. The result \( r=A/2 \) uses only the prism angle and the symmetry at minimum deviation. The refractive index does not appear in it. So each colour has \( r_1=30^\circ \).

Option (2) greater for the violet colour — Incorrect. A different refractive index changes the angle of incidence at which violet light reaches minimum deviation. The angle of refraction inside the prism stays \( A/2 \).

Option (3) greater for the red colour — Incorrect. The colours differ in the angle of minimum deviation \( D_m \), through \( n=\sin[(A+D_m)/2]/\sin(A/2) \). But \( r_1 \) stays \( A/2 \) for red as well.

Option (4) equal for both the colours, but not \( 30^\circ \) — Incorrect. The angles are equal, but their common value is fixed at half the prism angle, which is \( 30^\circ \) for a \( 60^\circ \) prism.

NCERT (2026-27):
Refraction Through a Prism – “Comparing these two equations, we get

    \( r_1+r_2=A \)\( \qquad (9.34) \)

… At the minimum deviation D_m, the refracted ray inside the prism becomes parallel to its base. We have δ = D_m, i = e which implies r₁ = r₂.”