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
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 .
For important questions based on the chapter wave optics– click here