Interference of light-

Definition: The phenomenon of redistribution of light energy due to the superposition of two coherent light waves of the same frequency (or wavelength), producing alternate bright and dark fringes, is called interference of light.

What are the coherent sources of light?

Explanation

When two coherent light waves travelling through the same medium superimpose on each other, the resultant intensity at different points changes due to constructive and destructive interference.

Expression for the Conditions of Constructive and Destructive Interference

When two coherent light waves superimpose, the resultant intensity depends on the path difference or phase difference between them.

Let the two wave

Y₁=A₁ sin (ωt )

Y₂=A₂ sin (ωt +Φ)

After superimposition

 Y = Y₁ + Y₂

Y = A₁ sin (ωt )+ A₂ sin (ωt +Φ)

After solving it we get ,

Y = R A₂ sin (ωt +θ)

Where R is the resultant amplitude of the resulting wave

R² = A₁ ² + A₂² +2 A₁A₂ cos (ωt )

But Intensity of light,  I α (amplitude )²

So we can write ,

I = I₁ + I₂ +2√I₁I₂ cos ωt ,

1. Constructive Interference (Bright Fringe)

When two waves meet in the same phase, their amplitudes add up and maximum intensity is obtained.

 For Maximum intensity cos Φ = +1

cos Φ = 2nπ

 Therefore path difference  p = nʎ [ whwre n= 0,1,2,3…],

For maximum intensity I_max = I₁ + I₂ +2√I₁I₂.

Or, I_max = A₁²+ A₂ ² + 2A₁ A₂ = (A₁ +A₂)²

2. Destructive Interference (Dark Fringe)

When two waves meet in opposite phases, their amplitudes cancel each other and minimum intensity is obtained.

Condition: intensity should be minimum i.e. cos Φ = -1

 Phase difference , Φ = (2n-1)π

Path difference p = (2n-1 )ʎ/2

For minimum intensity I_min = I₁ + I₂ – 2√I₁I₂,

OR, I_m = A₁²+ A₂ ² – 2A₁ A₂ = (A₁ -A₂)²

Diffraction of light

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Objective questions of wave optics

Diffraction of light

What is diffraction of light ? conditions for maximum and minimum using single slit experiment, and intensity curve .

Diffraction of light  –

When a beam of monochromatic light passes through an aperture or an obstacle comes into the way of light then light is bent at the corners of the aperture or obstacle , this phenomena of light is called diffraction of light

Diffraction of light at a single slit – when a beam of monochromatic light is made incident on single slit , then we get  diffraction pattern on the screen , which contain bright (maximum) and dark (minimum) band .

 

Let, AB be a slit of width ‘a’ and parallel beam of a monochromatic light incident on it . All the primary ray incident converges at the center and form a bright fringe, and all the secondary wave gets superimposed and form secondary dark and bright fringes on the screen .

Let θ be the angle of diffraction  for waves reaching at point P of screen , and AD is the perpendicular drown from ‘A’

The path difference   between the rays AP and BP

So path difference , p =   BP – AP = BD

But from geometry <ANB = θ

Sinθ= BD/AB ,

Or,   BD = AB sinθ

width of the slit AB = a

Hence path difference ;  p = a sinθ  ……………………………(i)

Position of central maximum  – the wavelets fall on the lens L along parallel direction of the axis reaches at o in the same phase and form central maximum .

Position of secondary minimum  –   if we imagine the slits is divided in two equal halves  AC and CB then every point on the upper half the path difference is λ/2  and same as in lower half . they will reach at the point p in opposite phase and form destructive  interference cause of minimum or dark band , for dark band   a sinθ =  nλ  , where  n= 1,2,3,4………..

, if θ is very small then , aθ = nλ

So , θ = nλ/a  .

Position of secondary maximum  –   we imagine slits is divided in three equal parts so the path difference between these two upper parts will be λ/2 , which will cancelled each other effect , so only third part will be the cause of the diffraction pattern obtained on the screen .

So ,    a sinθ    = ( 2n+1) λ/2

If     θ is very small then ,      a θ  =    ( 2n+1) λ/2

So ,            θ =( 2n+1) λ/2  / a

 

Intensity- position curve for diffraction of light

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Central maximum is twice the width of any other secondary maximum –

 

For first minimum , from the relation of secondary minimum     θ = nλ/a

Here value of n=1 ( for first minimum ) , so,  θ = 1λ/a  .

But  , according to the geometry  θ = y₁ / D ( where D is the separation between the slit and screen ) .

Angular width of the central maximum           2θ = y₁ + y₂  /D ,

Or,          2λ/a =  y₁+ y₂ /D,

Or ,     Y₁ + y₂ = 2λD/a

Therefore , width of the slits = 2( λD/a) = 2  x (width of other secondary maximum).

Questions based on diffraction of light  – What do you mean by the term diffraction of light ? using single slit show the conditions of central maximum, secondary minimum, secondary maximum and hence draw the intensity curve with position ‘x’ . Using single slit experiment show that (i) the intensity of diffraction fringes increases as order (n) increases .(ii) angular width of the central maximum is twice of first order secondary maximum .

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