In the Huygens construction for reflection, a plane wavefront AB meets a plane mirror first at A. In the time the edge B takes to reach the mirror at C, a secondary wavelet of radius AE grows from A. The tangent CE to this wavelet is the reflected wavefront. The triangles EAC and BAC are congruent because
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → AC is common, AE = BC, and the angles at B and E are right angles
The construction rests on time. The wavelet from A and the edge B both travel for the same time \( \tau \) in the same medium, at the same speed \( v \).
Option (1) AC is common, and AE = 2BC because the wavelet goes to the mirror and back — Incorrect. The wavelet from A grows only for the time \( \tau \). So its radius is \( v\tau \), equal to BC, with no factor 2.
Option (2) AC is common, AE = BC, and the angles at A and C are right angles — Incorrect. The angles at A and C are not right angles. They depend on the angle of incidence. The right angles are at B, where the ray meets the wavefront AB, and at E, where the tangent CE meets the radius AE.
Option (3) AC is common, and AE is less than BC because the reflected wave moves more slowly — Incorrect. The incident and reflected waves are in the same medium. They have the same speed, so AE equals BC.
Option (4) AC is common, AE = BC, and the angles at B and E are right angles — Correct. Here AE and BC are both equal to \( v\tau \), and AC is a shared side. The ray BC is perpendicular to the wavefront AB, and the tangent CE is perpendicular to the radius AE. So the two right triangles are congruent, and \( i=r \).
NCERT (2026-27):
Reflection of a plane wave by a plane surface – “In order to construct the reflected wavefront we draw a sphere of radius vτ from the point A … Let CE represent the tangent plane drawn from the point C to this sphere. Obviously AE = BC = vτ”
Summary – “… The rays are then perpendicular to both the wavefronts and the time of travel is the same measured along any ray. …”