Name: Waves and Their Phases
Uploaded: 2019-07-19
Duration: 6 min 51 s
Description: In this video, we will learn how two or more waves can be studied in terms of their phases. Equations of two waves and graphs of two waves. Learn more!

Question

Describe diffraction of light due to a single slit. Explain formation of a pattern of fringes obtained on the screen and plot showing variation of intensity with angle

θ

in single slit diffraction.

Accepted Answer

Diffraction of light at a single slit: When monochromatic light is made incident on a single slit. we get diffraction pattern on a screen placed behind the slit. The diffraction pattern contains bright and dark bands. the intensity of central band is maximum and goes on decreasing on both sides.
Explanation: Let AB be a slit of width $a$ and a parallel beam of monochromatic light is incident on it. According to Fresnel the diffraction pattern is the result of superposition of a large number of waves. starting from different points of illuminated slit.
Let $θ$ be the angle of diffraction for waves reaching at point $P$ of screen and $A N$ the perpendicular dropped from A on wave diffracted from B.
The path difference between rays diffracted at points A and B.

$Δ = B P - A P = B N In Δ A N B, ∠ A N B = 90 ° and ∠ B A N = θ$

$\sin θ = \frac{B N}{A B} or B N = A B \sin θ$

As AB = Width of slit = $a$

$P a t h d i f f e r e n c e, Δ=asinθ$

To find the effect of all coherent waves at $P$ . We have to sum up their contribution. each with a different phase. This was done by Fresnel by rigorous calculations, but the main features may be explained by simple arguments given below:
At the central point C of the screen, the angle $θ$ is zero. Hence the waves starting from all points of slit arrive in the same phase. This gives maximum intensity at the central point C.
Minima: Now we divide the slit into two equal halves AO and OA each of width $2 a$ . Now for every point, M1 in AO, there is a corresponding point M2 in OB, such that

$M_{1} M_{2} = \frac{a}{2}$ ; then path difference between waves arriving at P and starting from M1 and M2 will be $2 a \sin θ = 2 λ $ . This means that the contributions from the two halves of slit AO and OB are opposite in phase and so cancel each other. It gives the angle of diffraction at which intensity falls to zero. Similarly it may be shown that the intensity is zero for $\sin (θ) \frac{n λ}{a}$ .

Thus the general condition of minima is $a \sin θ = n λ$

Secondary Maxima: Let us now consider angle θ such that, $\sin (θ) = θ = \frac{3 λ}{2 a}$

Which is midway between two dark bands given by, $\sin θ = θ = \frac{λ}{a}$ $and \sin θ = θ = \frac{2 λ}{a}$

Clearly there will be a maxima between first two minima. but this maxima will be of much weaker intensity than central maximum. This is called first secondary maxima. In a similar manner we can show that there are secondary maxima between any two consecutive minima: and the intensity of maxima will go on decreasing with increase of order of maxima. In general the position of nth maxima will be given by

$a \sin θ = (n + \frac{1}{2}) λ$

The intensity of secondary maxima decreases with increase of order n because with increasing n. the contribution of slit decreases.

Describe diffraction of light due to a single slit. Explain formation of a pattern of fringes obtained on the screen and plot showing variation of intensity with angle $θ$ in single slit diffraction.

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