A ray of light is incident on one face of an equilateral prism at an angle of incidence of \( 50^\circ \). At this angle the prism is in the position of minimum deviation. The angle of minimum deviation is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 40^\circ \)
For a prism the angle of deviation is \( \delta=i+e-A \). At minimum deviation the ray passes symmetrically through the prism, so the angle of emergence equals the angle of incidence, \( e=i \). An equilateral prism has \( A=60^\circ \).
\( \begin{aligned} D_m &= 2i-A \\ &= 2\times50^\circ-60^\circ \\ &= 40^\circ \end{aligned} \)
NCERT (2026-27):
Refraction Through a Prism – “… The total deviation δ is the sum of deviations at the two faces,
\( \delta=(i-r_1)+(e-r_2) \)
that is,
\( \delta=i+e-A \qquad (9.35) \)
… At the minimum deviation Dm, the refracted ray inside the prism becomes parallel to its base. We have δ = Dm, i = e which implies r1 = r2.”