Question
Using Huygens principle, explain reflection of a plane wave by a plane reflecting surface.
✓
Answer
→ Consider a plane wave AB incident at an angle $i$ on a reflecting surface MN .
→ If $u$ is the speed of wave in the given medium, $\tau$ represents the time taken by the wavefront to advance from point $B$ to $C$, then the distance $BC =v \tau$.
→ In order to construct the reflected wavefront we draw a sphere of radius $v$ from the point $A$ as shown in fig.
→ Let CE represent the tangent plane drawn from the point C to this sphere.
→ Obviously,
$
AE=BC=UT
$
→ From fig., incident and reflected wave fronts make angle $i$ and $r$ with reflecting surface, MN respectively.
→ From fig.,
AC is the common side between $\triangle AEC$ and $\triangle ABC$,
Also, $\angle AEC =\angle ABC =\frac{\pi}{2}$
$
\text { and } AE=BC=UT
$
→ So $\triangle AEC$ and $\triangle ABC$ are congruent.
→ And therfore, $\angle i=\angle r$,
which is law of reflection.
→ If $u$ is the speed of wave in the given medium, $\tau$ represents the time taken by the wavefront to advance from point $B$ to $C$, then the distance $BC =v \tau$.
→ In order to construct the reflected wavefront we draw a sphere of radius $v$ from the point $A$ as shown in fig.
→ Let CE represent the tangent plane drawn from the point C to this sphere.
→ Obviously,
$
AE=BC=UT
$
→ From fig., incident and reflected wave fronts make angle $i$ and $r$ with reflecting surface, MN respectively.
→ From fig.,
AC is the common side between $\triangle AEC$ and $\triangle ABC$,
Also, $\angle AEC =\angle ABC =\frac{\pi}{2}$
$
\text { and } AE=BC=UT
$
→ So $\triangle AEC$ and $\triangle ABC$ are congruent.
→ And therfore, $\angle i=\angle r$,
which is law of reflection.
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