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 affect the size and intensity of the central diffraction bond? 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 :

A wavefront is defined as the continuous locus of all the particles which are vibrating in the same phase.

The perpendicular line drawn at any point on the wavefront represents the direction of propagation of the wave at that point.

Verification of Laws of Reflection using Huygen's Principle

In the given figure, AB is the wavefront incident on a reflecting surface XY with an angle of incidence i as shown in figure. According to Huygen’s principle, every point on AB acts as a source of secondary wavelets. At first, wave incidents at point A and then to points C, D and E. They form a sphere of radii AA1, CC1 and DD1 as shown in figure.

A1E represents the tangential envelope of the secondary wavelet in forward direction.

In ΔABE and ΔAA1E,

∠ABE = ∠AA1E = 90°

Side AE = Side AE

AA1 = BE = distance travelled by wave in same time

So, these triangles are congruent.

So, ∠BAE = i and ∠BEA =r

Thus, i = r

Hence, angle of incidence is equal to the angle of reflection. This is the first law of reflection.

Since, the incident wavefront AB, normal and reflected wavefront A1E lies in same plane, it verifies the second law of reflection.

In this way, laws of reflection is verified by Huygens’s principle.

(b) The width of the central maximum is given by $\frac{2D\lambda }{d}$ where d is the slit width.

On doubling the width of the slit, the size of diffraction band reduces to half value.

When the width of central maxima is reduced to half, area of central diffraction band will become one fourth. And hence the intensity will become four times.

(c) It is so because wave diffracted from the edge of the circular obstacle interfere to form constructive interference at the centre of the obstacle.