Q.Derive the laws of reflection of light using Huygens' principle.

[!TLDR] Congruent triangles formed by the Huygens wavelet from the first-struck point and the geometry of the last-struck point give angle of incidence = angle of reflection directly, with both rays and the normal shown to lie in one plane. [!ANSWER] Using Huygens' construction, congruent triangles ABC and AEC (sharing hypotenuse AC, with $AE=BC=vT$) give $\angle ACE=\angle BAC=i$ and $\angle ACE=\angle PAE=r$, so $i=r$ -- the law of reflection -- with the incident ray, normal and reflected ray all lying in the plane of incidence.

Consider a plane wavefront AB, perpendicular to the plane of the page, incident at angle i (measured from the normal) on a plane mirror MN, also perpendicular to the page. Point A of the wavefront reaches the mirror first, at time $t=0$; point B, further along the wavefront, reaches the mirror only later, at a point C on the mirror, at time $t=T$. Between these two instants, every point along AB between A and C strikes the mirror in succession and, by Huygens' Principle, immediately begins emitting its own secondary (reflected) hemispherical wavelet, all on the SAME side of the mirror as the incident wave (since reflection keeps the light in the original medium).

By the time T when B finally reaches the mirror (at C), the wavelet that started from A (the very first point struck) has grown to its full radius $AE=vT$; the wavelet from C itself (only just struck, at time T) has radius zero; wavelets from points struck in between have intermediate, progressively smaller radii. The REFLECTED wavefront at time T is the common tangent (envelope) to this entire family of wavelets -- the straight line EC, tangent to the full wavelet centred at A.

Since A's wavelet radius $AE=vT$ exactly equals $BC=vT$ (the distance light travels in the SAME medium, at the SAME speed v, in the SAME time T), the two right-angled triangles ABC and AEC share a common hypotenuse AC and have one pair of equal sides ($AE=BC$) -- by the RHS (right angle-hypotenuse-side) congruence criterion, the two triangles are CONGRUENT. From this congruence, $\angle ACE=\angle BAC$ (angles opposite the equal sides in congruent triangles). Now, $\angle BAC$ is exactly the angle of incidence i: ray RA is perpendicular to the incident wavefront AB, and the normal AP is perpendicular to the mirror MN, so the angle between ray RA and normal AP (which is i, by definition) equals the angle between wavefront AB and line AC -- which is $\angle BAC$. Similarly, since the reflected ray AE is perpendicular to the reflected wavefront EC, and the normal AP is perpendicular to AC, the angle $\angle ACE$ equals $\angle PAE$, which is exactly the angle of reflection r (the angle between the reflected ray AE and the normal AP). Combining the congruence result $\angle ACE=\angle BAC=i$ with $\angle ACE=\angle PAE=r$ gives directly $i=r$ -- the FIRST law of reflection.

It is also immediately visible from this construction that the incident ray RA, the normal AP, and the reflected ray AE all lie within the single plane of the page (the plane containing the original wavefront's direction of travel and the normal) -- this is the SECOND law of reflection, following automatically from the fact that the whole Huygens' construction was carried out entirely within that one plane, with no component of the wavelets' geometry ever leaving it. [!ANSWER] Using Huygens' construction, congruent triangles ABC and AEC (sharing hypotenuse AC, with $AE=BC=vT$) give $\angle ACE=\angle BAC=i$ and $\angle ACE=\angle PAE=r$, so $i=r$ -- the law of reflection -- with the incident ray, normal and reflected ray all lying in the plane of incidence.

Construct the Huygens secondary wavelet from the first point of the wavefront to strike the mirror, identify the reflected wavefront as its common tangent with the last-struck point, and use the resulting congruent right triangles (equal hypotenuse, equal side $vT$) to prove $i=r$ geometrically.

Forgetting to also state and justify the SECOND law of reflection (all three -- incident ray, normal, reflected ray -- lying in one plane), treating the derivation as complete once $i=r$ alone has been shown.