Subject: Applied Physics 2

Topic: Interference And Diffraction Of Light

Difficulty: High

: Consider a beam of light incident on two narrow slits. Let ‘a’ be the width of slit an ‘b’ the width of opaque interval. The diffraction pattern is found to consist of equally spaced interference maxima and minima in the region originally occupied by the central maxima in the single slit.

Let us imagine that all the waves from the single slit is replaced by a single wave starting from the centre of amplitude $E_m$ sinα / α .

The path difference between them is (a+b) sinθ =λ

For path difference of λ phase difference is 2π

Therefore for path difference of (a+b) sinθ , phase difference is Δ∅ = (2П(a+b) sinθ) / λ

let Δ∅ =2ꞵ where ꞵ = {П(a+b)} sinθ / λ.

The intensity distribution for N slits is given by

$I_θ$ = $I_m$ ($sin^2$ α ) / $α^2$ ($sin^2$ Nβ) / ($sin^2$ β)

For double slit Put N =2,

Sin 2ꞵ = 2sinꞵ cosꞵ

Therefore, $I_θ$ = [$I_m$ ($sin^2$ α ) / $α^2$ ] ($4sin^2$ β$cos^2$ β)/($Sin^2$ β)

$I_θ$ = [ 4 $I_m$ ($sin^2$ α ) / $α^2$] $cos^2$ ꞵ

The intensity will be maximum for double slit If $Cos^2$ ꞵ =1

⸫ ꞵ =±mП

(П(a+b) sinθ ) / λ. =±mП

⸫ (a+b) sinθ =mλ…………………………….equation for maxima.

The intensity will be minimum if $cos^2$ ꞵ =0

i.e ꞵ = (2n± 1) П /2

{П(a+b) sinθ } / λ= (2n± 1) П /2

(a+b) sinθ = (2n± 1)λ /2………… equation of minima

The entire pattern due to double slit may be considered as consisting of interference fringes due to light from both slits. The intensities of these fringes being governed by diffraction occurring at individual slits.