Question

Write Huygen's principle of secondary wavelets and verify laws of refraction on its basis.
OR
Explain the meaning of coherent sources and sustained interference. Why central fringe in Young's double slit experiment is always bright? In Young's double slit experiment, fringe width is 0.04 cm for light of 6000 Å. Distance of screen from slits is 1 m. Find the distance of third bright and second dark fringe from central fringe and also the separation between the slits.

Hint:

Use Huygens' wavelet envelope: \(\sin i/\sin r=v_1/v_2\). For the numerical, \(\beta=\lambda D/d\), bright fringe distance \(=n\beta\), dark fringe distance \(=(2n-1)\beta/2\).

Answer 1

Option 1: Huygens' principle and laws of refraction

Step 1 (Principle): Every point on a given wavefront acts as a fresh source of disturbance, called a secondary wavelet, which spreads out in all forward directions with the speed of the wave in that medium. The new wavefront at a later time is the forward tangential envelope (common tangent surface) of all these secondary wavelets.

Step 2 (Set-up for refraction): Let a plane wavefront \(AB\) travel in medium 1 (speed \(v_1\)) and strike a plane boundary \(XY\) at an angle of incidence \(i\). While the wavelet from \(B\) travels to \(C\) on the surface in time \(t\), the wavelet from \(A\) enters medium 2 (speed \(v_2\)) and spreads a distance \(AD\).

Step 3 (Distances): In time \(t\):
\[ BC=v_1 t \quad\text{and}\quad AD=v_2 t. \]
Drawing the tangent from \(C\) to the sphere of radius \(AD\) about \(A\) gives the refracted wavefront \(CD\), making angle of refraction \(r\) with the surface.

Step 4 (Geometry): In right triangle \(ABC\), \(\sin i=\dfrac{BC}{AC}=\dfrac{v_1 t}{AC}\). In right triangle \(ADC\), \(\sin r=\dfrac{AD}{AC}=\dfrac{v_2 t}{AC}\).

Step 5 (Snell's law): Dividing,
\[ \frac{\sin i}{\sin r}=\frac{v_1}{v_2}=\frac{c/v_2}{c/v_1}=\frac{n_2}{n_1}=\text{constant}. \]
This is Snell's law, proving the ratio \(\sin i/\sin r\) is constant (first law). Also the incident ray, refracted ray and normal all lie in the plane of the paper (plane of incidence), which is the second law. Hence both laws of refraction are verified.
\[ \boxed{\;n_1\sin i=n_2\sin r\;} \]

Option 2: Coherent sources, sustained interference and the numerical

Step 1 (Coherent sources): Two sources are coherent if they emit waves of the same frequency (and wavelength) with a constant phase difference that does not change with time. In practice they are derived from a single source (as the two slits in YDSE).

Step 2 (Sustained interference): Sustained (stable) interference means the positions of bright and dark fringes stay fixed in space and time, giving a steady pattern. This requires coherent sources of equal (or nearly equal) amplitude and small slit separation.

Step 3 (Central fringe bright): At the central point, both waves travel equal distances from the two slits, so the path difference is zero \((\Delta x=0)\) and the phase difference is zero for all wavelengths. Constructive interference is therefore complete, making the central fringe always bright (and white in white light).

Step 4 (Data): Fringe width \(\beta=0.04\text{ cm}=4\times10^{-4}\text{ m}\), \(\lambda=6000\ \text{\AA}=6\times10^{-7}\text{ m}\), \(D=1\text{ m}\).

Step 5 (Slit separation): Since \(\beta=\dfrac{\lambda D}{d}\),
\[ d=\frac{\lambda D}{\beta}=\frac{6\times10^{-7}\times1}{4\times10^{-4}}=1.5\times10^{-3}\text{ m}=1.5\text{ mm}. \]
Step 6 (Third bright fringe): Distance of \(n\)-th bright fringe \(=n\beta\). For \(n=3\):
\[ y_3=3\beta=3\times0.04=0.12\text{ cm}. \]
Step 7 (Second dark fringe): Distance of \(n\)-th dark fringe \(=\dfrac{(2n-1)}{2}\beta\). For \(n=2\):
\[ y_{2}^{\,dark}=\frac{(2\times2-1)}{2}\beta=\frac{3}{2}\times0.04=0.06\text{ cm}. \]
\[ \boxed{\;d=1.5\text{ mm},\;\; y_{3\,bright}=0.12\text{ cm},\;\; y_{2\,dark}=0.06\text{ cm}\;} \]