Question

(a) Define a wavefront. Using Huygen’s principle, verify the laws of reflection at a plane surface.

(b) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this effect the size and intensity of the central differaction band? Explain.

(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the obstacle. Explain why.  

Answer 1

(a) Wavefront is the locus of the particles of the medium in same vibrating phase.

Laws of reflection at a plane surface

Let XY be a plane reflecting surface and AB be a plane wavefront incident on the surface as shown in Fig. 

According to Huygens’ principle, every point on wavefront AB is a source of secondary wavelets and the time during which wavelet from B reaches at C, the reflected wavelet from A would arrive at D.

If t is the time taken by wavelet from B to C [or A to D] then

Putting Eq. (ii) and (iii) in Eq. (i), we get

AC sin i = AC sin r

or sin i = sin r

or i = r

i.e. Angle of incidence = Angle of reflection.

This proves first law of reflection.

Second law of reflection

Since incident ray, reflected ray and the normal all lie on the same plane i.e. plane XY, so second law of reflection is also proved.

(b) Size of diffraction pattern

y₀ = 2\(\frac{\lambda D}{d}\)

When the slit is made double the original width i.e. d = 2d

y = \(\frac{2\lambda D}{d} = \frac{1}{2}(\frac{2\lambda D}{d}) = \frac{y_0}{2}\) 

Hence the area of the central diffraction pattern becomes \(\frac{1}{4}\). So, the intensity of central diffraction pattern becomes 4-times.

(c) When circular obstacle is placed in the path of light as shown in the given figure, then the waves get different from the edges of the circular obstacle and interfere constructively at the centre and hence a bright spot is seen at the centre.