Question

Define Wavefront. Explain, refraction of waves with the help of Huygen's secondary wavelet principle.

Hint:

The key to proving laws of reflection or refraction using Huygen's principle is to correctly identify the distances traveled by the wavelets from two different points on the incident wavefront in the same interval of time. The new wavefront is the common tangent to all these secondary wavelets.

Answer 1

1. Definition of Wavefront:
A wavefront is defined as the continuous locus of all the particles of a medium which are vibrating in the same phase at any given instant. The direction of propagation of the wave is always perpendicular to the wavefront.
2. Refraction of Waves using Huygen's Principle:
Let us consider a plane wavefront AB incident on a refracting surface XY, separating two media, medium 1 (rarer) and medium 2 (denser). Let the speed of light in the two media be \(v_1\) and \(v_2\) respectively, with \(v_1 > v_2\). The angle of incidence is \(i\). \begin{center} \begin{tikzpicture}[scale=1.5] % Interface \draw[thick] (-2,0) -- (3,0) node[right] {Y}; \node at (-2,0) [left] {X}; \node at (0, 1.5) {Medium 1 (speed $v_1$)}; \node at (0, -1) {Medium 2 (speed $v_2$)}; % Incident rays and wavefront \draw[->, thick, blue] (-1, 2) -- (0,0); \draw[->, thick, blue] (1.5, 2) -- (2.25, 0); \draw[thick, red] (-1,2) node[above left] {A} -- (1.5,2) node[above right] {B}; % Normals \draw[dashed] (0,0) -- (0,-2); \draw[dashed] (2.25,0) -- (2.25,-2); % Angles of incidence \draw (0,0.8) arc (90:230:0.8); \node at (-0.3, 0.5) {$i$}; \draw (2.25,0.8) arc (90:230:0.8); \node at (1.95, 0.5) {$i$}; % Construction \node at (2.25,0) [above] {C}; \node at (0,0) [above] {A'}; % Time calculation % Let time for B to reach C be t. BC = v1*t. % BC is hypotenuse of triangle ABC. AC = 2.25 - (-1) = 3.25. Angle is i. % sin(i) = BC/AC'. A' is (0,2). C is (2.25,0). No, this is complex. % Let's use the standard way. t = BC/v1. In time t, wavelet from A' travels AD = v2*t. % Draw the refracted wavefront % BC = A'C * sin(i). Let A' be origin for a moment. No. Let A' be (0,0). % B is at (some x, some y). This is too complex. Let's draw it schematically. \draw[red] (1.5,2) -- (2.25,0); % Ray BC \node at (1.7, 1.1) {$v_1 t$}; % Secondary wavelet from A' % AD = v_2*t = v_2 * (BC/v_1) \draw[dashed, blue] (0,0) circle (1.2); % AD = v_2*t \node at (0.9, -0.9) [below right] {D}; % Refracted wavefront \draw[thick, red] (2.25, 0) -- (0.9, -0.9); % Tangent CD % Refracted ray \draw[->, thick, blue] (0,0) -- (0.9, -0.9); % Angle of refraction \draw (0, -0.5) arc (270:210:0.5); \node at (0.2, -0.4) {$r$}; \end{tikzpicture} \end{center} \begin{itemize} \item According to Huygen's principle, every point on the incident wavefront AB is a source of secondary wavelets. \item When the wavefront strikes the surface at point A, A becomes a source of secondary wavelets in the second medium. Meanwhile, point B is still in the first medium. \item Let \(t\) be the time taken for the disturbance from point B to reach point C on the surface. Therefore, the distance \(BC = v_1 t\). \item In the same time \(t\), the secondary wavelets from point A will travel a distance \(AD = v_2 t\) in the second medium. \item To find the new refracted wavefront, we draw a sphere (a circle in 2D) of radius \(AD = v_2 t\) with A as the center. \item The tangent drawn from point C to this sphere gives the new refracted wavefront, CD. \item Now, in the right-angled triangle ABC: \[ \sin i = \frac{BC}{AC} = \frac{v_1 t}{AC} \] \item And in the right-angled triangle ADC: \[ \sin r = \frac{AD}{AC} = \frac{v_2 t}{AC} \] \item Dividing the two equations, we get: \[ \frac{\sin i}{\sin r} = \frac{v_1 t / AC}{v_2 t / AC} = \frac{v_1}{v_2} \] \item The ratio of the speeds of light is equal to the refractive index of the second medium with respect to the first, \(n_{21}\). \[ \frac{\sin i}{\sin r} = \frac{v_1}{v_2} = n_{21} \] \end{itemize} This is Snell's law of refraction, thus proving the law using Huygen's principle.