A ray of light falls at an angle of incidence of \( 70^\circ \) on one face of a prism of angle \( 60^\circ \) and comes out of the other face at an angle of emergence of \( 40^\circ \). For which other angle of incidence on the same face will the ray suffer the same deviation?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 40^\circ \)
The deviation produced by a prism is \( \delta=i+e-A \). This relation does not change when \( i \) and \( e \) are interchanged, and the path of a ray through the prism can be traced back. So a ray sent in at the old angle of emergence comes out at the old angle of incidence. It has the same deviation.
The deviation in the given case is:
\( \begin{aligned} \delta &= i+e-A \\ &= 70^\circ+40^\circ-60^\circ \\ &= 50^\circ \end{aligned} \)
With the angles interchanged, \( i'=40^\circ \) and \( e'=70^\circ \):
\( \begin{aligned} \delta' &= 40^\circ+70^\circ-60^\circ \\ &= 50^\circ \end{aligned} \)
Therefore, an angle of incidence of \( 40^\circ \) gives the same deviation of \( 50^\circ \). The ray then leaves the other face at \( 70^\circ \). Only at minimum deviation, where \( i=e \), is there a single angle of incidence for a given deviation.
NCERT (2026-27):
Refraction Through a Prism – “… The total deviation δ is the sum of deviations at the two faces, δ = (i – r1) + (e – r2) that is,”
\( \delta=i+e-A \qquad (9.35) \)
“… in general, any given value of δ, except for i = e, corresponds to two values i and hence of e. This, in fact, is expected from the symmetry of i and e in Eq. (9.35), i.e., δ remains the same if i and e are interchanged.”