A ray of light passes through a prism of angle \(A\). The angle of refraction at the first face is \(r_1\) and the angle of incidence at the second face is \(r_2\). The normals to the two faces meet at a point N. The quadrilateral formed by the two faces and the two normals has two right angles. Comparing it with the triangle formed by the ray inside the prism and the two normals gives
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(r_1+r_2=A\)
The angles of a quadrilateral add up to \(360^\circ\). Two of them are \(90^\circ\), so the other two add up to \(180^\circ\). The angles of the triangle also add up to \(180^\circ\).
\( \begin{aligned} \angle A+\angle N &= 180^\circ \\ r_1+r_2+\angle N &= 180^\circ \end{aligned} \)
\( r_1+r_2=A \)
Option (1) \(i+e=A\) — Incorrect. The angles \( i \) and \( e \) lie outside the prism. They appear in the deviation relation \( \delta=i+e-A \).
Option (2) \(r_1+r_2=A\) — Correct. Both sums equal \(180^\circ\) and share \( \angle N \). So \( r_1+r_2=A \) for every ray.
Option (3) \(r_1+r_2=180^\circ-A\) — Incorrect. The angle \(180^\circ-A\) is the angle at N between the two normals. The sum \( r_1+r_2 \) equals \( A \) itself.
Option (4) \(r_1=r_2=A/2\) — Incorrect. This holds only at minimum deviation, where \( r_1=r_2 \). For a general ray, \( r_1 \) and \( r_2 \) are different.
NCERT (2026-27):
Refraction Through a Prism – “In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180°.”
\( \displaystyle \angle A+\angle QNR=180^\circ \)
“From the triangle QNR,”
\( \displaystyle r_1+r_2+\angle QNR=180^\circ \)
“Comparing these two equations, we get”
\( \displaystyle r_1+r_2=A \)\( \qquad (9.34) \)