Question

(a) Consider two coherent sources $S_1$ and $S_2$ producing monochromatic waves to produce interference pattern. Let the displacement of the wave produced by $S_1$ be given by $Y_1=acos\omega t$ and the displacement by $S_2$ be $Y_2=acos(\omega t+\varphi )$. Find out the expression for the amplitude of the resultant displacement at a point and show that the intensity at that point will be $I=4a^{2}co{s}^2\frac{\varphi }{2}$. Hence, establish the conditions for constructive and destructive interference.
(b) What is the effect on the interference fringes in Young's double slit experiment when (i) the width of the source slits is increased; (ii) the monochromatic source is replaced by a source of white light.

Answer 1

(a) Let S be a narrow slit illuminated by a monochromatic source of light and $S_1$ and $S_2$ two similar parallel slits very close together and equidistant from S.



Displacement of the wave produced by $S_1$ is given by

$y_1=acos\omega t$

and the displacement of the wave produced by $S_2$ is given by

$y_2=acos(\omega t+\varphi )$

The resultant displacement is given by

$y=y_1+y_2$
$y=acos\omega t+acos(\omega t+\varphi )$
$y=2acos\varphi /2cos(\omega t+\varphi /2)$

The amplitude of the resultant displacement is $2acos\varphi /2$. The intensity of light is directly proportional to the square of amplitude of the wave.

The resultant intensity is given by

$I=4a^{2}co{s}^2\varphi /2$

For constructive interference the intensity of light at point P is maximum, if

$cos\varphi =max=+1:\varphi =0,2\pi ,4\pi ....$

$\varphi =2n\pi$ where n=0, 1, 2, .....

Path difference $=\frac{\lambda }{2\pi }\times 2n\pi =n\lambda$

For distructive interference the intensity of light at point P is minimum, if

$cos\varphi =-1$

$\therefore \varphi =\pi ,3\pi ,5\pi ,.....$

or $\varphi =(2n-1)\pi$ where n=1, 2, 3, ....

The corresponding path difference between the two waves

$x=\frac{\lambda }{2\pi }\varphi =\frac{\lambda }{2\pi }(2n-1)\pi =(2n-1)\frac{\lambda }{2}$

(b) (i) As the width of the slits is increased, the fringe width decrease. It is because,

$\beta \propto \frac{1}{d}$

(ii) The different colours of white light will produce different interference patterns but the central bright fringes due to all colours are at the same positions. Therefore, the central bright fringe is white in colour. Since the wavelength of the blue light is smallest, the fringe closed on the either side of the central white fringe is blue and farthest is red. Beyond a few fringes, no clear fringe pattern will be visible.