A ray of light falls at an angle of incidence \(i\) on one face of a rectangular glass slab kept in air. It leaves from the opposite parallel face. The angle of refraction \(r_1\) at the first face equals the angle of incidence at the second face. Snell’s law at the two faces shows that the emergent ray
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → is parallel to the incident ray but shifted sideways
Snell’s law links the angles on the two sides of each face. The faces are parallel, so the ray meets both faces at the same angle \( r_1 \) inside the glass. Here \( e \) is the angle of emergence.
\( \begin{aligned} \sin i &= n\sin r_1 \\ \sin e &= n\sin r_1 \end{aligned} \)
\( e=i \)
Option (1) bends towards the normal and leaves at an angle smaller than \(i\) — Incorrect. The ray bends towards the normal only inside the glass. At the second face it bends away again and leaves at the angle \( i \).
Option (2) comes out along the same line as the incident ray — Incorrect. Inside the slab the ray follows a slanting path. So the emergent ray is displaced sideways from the line of the incident ray.
Option (3) is parallel to the incident ray but shifted sideways — Correct. Since \( e=i \), the slab produces no deviation. It produces only a lateral shift.
Option (4) makes an angle \(i-r_1\) with the incident ray — Incorrect. The angle \( i-r_1 \) is the turn at the first face only. The second face turns the ray back by the same angle, so the net deviation is zero.
NCERT (2026-27):
Refraction – “… For a rectangular slab, refraction takes place at two interfaces (air-glass and glass-air). It is easily seen … that r2 = i1, i.e., the emergent ray is parallel to the incident ray—there is no deviation, but it does suffer lateral displacement/shift with respect to the incident ray.”