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(a) Describe any two characteristic features which distinguish between interference and diffraction phenomena. Derive the expression for the intensity at a point of the interference pattern in Young’s double slit experiment.

(b) In the diffraction due to a single slit experiment, the aperture of the slit is

$3\; mm$ . If monochromatic light of wavelength $620\; nm$ is incident normally on the slit, calculate the separation between the first order minima and the 3rd order maxima on one side of the screen. The distance between the slit and the screen is $1.5\; m$ .

Answers (1)

(a) Difference between Interference and Diffraction

Interference: A double-slit interference pattern has equally spaced dark and bright bands.
The peak intensity of the bright bands remains the same.
Diffraction: A single slit diffraction pattern has a central bright maximum that's twice as wide as other maxima.
The intensity falls as we go to the successive maxima away from the centre on either side.

Consider the two sources of light $S_{1}$ and $S_{2}$

Displacement produced by the source $S_{1}$ and $S_{2}$ can be written as,

$y_{1}=a_{0}\; \sin (wt)$

$y_{2}=a_{0}\; \sin (wt+\phi )$

Consider a point $P$ , on the screen.

At that point $(P)$ , by the principle of superposition, net displacement

$y=y_1+y_2$

$y=a_0sin\omega t+a_0sin(\omega t+\phi)$

$y=2a_{0}\; \cos \left ( \frac{\phi }{2} \right )\sin \left ( wt+\frac{\phi }{2} \right )$

We know that,

As intensity is the square of the amplitude.

$I=4\; a_{0}^{2}\; \cos ^{2}\left ( \frac{\phi }{2} \right )=4\; I_0\cos ^{2}\left ( \frac{\phi }{2} \right )$

(b) We have given the size of the aperture, $a\; =3mm=\; 3\times 10^{-3}m$

The wavelength of light, $\lambda =\; 620nm=\; 620\times 10^{-9}m$

Distance from the screen, $D=1.5 m$

The distance of first-order minima from the centre of central maxima

$=\frac{\lambda }{d}D=\frac{620\times 10^{-9}m}{3\times 10^{-3}m}\times 1.5=3.1\times 10^{-4}m$

The distance between first-order minima and the 3rd order maxima

$=2.5\frac{\lambda }{d}D= 7.75\times10^{-4}m$

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Safeer PP

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