Get Answers to all your Questions

header-bg qa

(a) What is a wavefront ? How does it propagate ? Using Huygens’ principle, explain reflection of a plane wavefront from a surface and verify the laws of reflection.

(b) A parallel beam of light of wavelength 500\; nm falls on a narrow slit and the resulting diffraction pattern is obtained on a screen
1\; m away. If the first minimum is formed at a distance of 2.5\; mm from the centre of the screen, find the (i) width of the slit, and
(ii) distance of first secondary maximum from the centre of the screen.

 

 

 

 
 
 
 
 

Answers (1)

a) A surface of constant phase is termed as a wavefront. The wave propagates in a direction perpendicular to the wavefront through secondary wavelets originating from
different points on it.

Let us consider a plane wave AB be incident on a reflecting surface and MN at an angle of incidence (i). Let \tau be the time taken by the wavefront to advance from B to C. Let v be the speed of the wave. Then,

BC\; = v\tau

Now, to draw the reflected wavefront. Let's draw a sphere of the radius centred at A. Now, In accordance with Huygen's principle, the tangent plane to this sphere passing through point c will give the refracted wavefront.

Remember that,

\angle BAC = i \; \; and\; \; \angle ACE = r

Let us consider the triangles EAC and BAC,

\because \angle AEC = \angle CBA

AE=BC=V\tau

AC = AC

So, by RHS, \Delta EAC\; \cong \Delta BCA

Thus, i = r

b) i) 

Given

The distance of the screen from the slit, D=1m

The distance of the first minimum  X_1=2.5mm=10^{-3}=2.5*10^{-3}mm

The wavelength of the light \lambda=500nm=500*10^{-9}m

Now,

As we know,

X_n=n\frac{\lambda D}{d}

d=n\frac{\lambda D}{X_n}=1*\frac{500*10^{-9}*1}{2.5*10^{-3}}=2*10^{-4}m=0.2mm

ii) 

For the first Secondary maxima

x=\frac{3\lambda D}{2d}=3.753mm

Posted by

Safeer PP

View full answer