The graph of the angle of deviation \(\delta\) against the angle of incidence \(i\) for a triangular prism has a minimum. A deviation a little larger than this minimum is produced at two angles of incidence, \(i_1\) and \(i_2\). Which of the following is correct?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → The angle of emergence for the ray incident at \(i_1\) is \(i_2\).
The deviation is \( \delta=i+e-A \). It stays the same if \( i \) and \( e \) are interchanged, because the path of the ray can be traced back. So each value of \( \delta \) above the minimum comes from two angles of incidence.
Option (1) The angle of emergence for the ray incident at \(i_1\) is \(i_2\). — Correct. The ray incident at \( i_1 \) leaves at \( e=i_2 \), and the ray incident at \( i_2 \) leaves at \( i_1 \). Both give the same \( \delta \).
Option (2) Both \(i_1\) and \(i_2\) lie on the same side of the minimum of the graph. — Incorrect. The curve falls to its minimum at \( i=e \) and then rises. One of the two angles is smaller than this value and the other is larger.
Option (3) The angle of refraction at the first face is the same for both rays. — Incorrect. The angles inside the prism are \( r_1, r_2 \) for one ray and \( r_2, r_1 \) for the other. They are equal only at minimum deviation, where \( r_1=r_2 \).
Option (4) The angle of emergence is the same for both rays. — Incorrect. The first ray leaves at \( i_2 \) and the second at \( i_1 \). These two angles are different.
NCERT (2026-27):
Refraction Through a Prism – “… You can see that, 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 …, i.e., δ remains the same if i and e are interchanged.”