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Fraunhofer Diffraction at a Single Slit

Let S is a source of monochromatic light of wavelength ‘A, L is collimating lens AB is a slit of width a, L’ is another conversing lens and XY is the screen light coming out from source and passing through slit is focused at the screen. A diffraction pattern is obtained on the screen which consists of central bright band having alternate dark and bright bands of decreasing intensity on both the sides. The complete arrangement is shown in Figure 5.6.

FIGURE 5.6 Fraunhofer’s diffraction at single slit

 

 

Analysis and explanation: According to Huygen’s theory a point in AB send out secondary waves in all directions. The diffracted ray along the direction of incident ray are focussed at C and those at an angle e and focussed at P and P’. Being at equidistant from all slits points, secondary wave will reach in same phase at C and so the intensity well be maximum. For the intensity at P, let AN is normal to BN, then path difference between the extreme rays is

 

∆ = BN = AN sin θ = a sin θ

= 2π/λ  a sinθ

 

which is zero for the ray from A and maximum for the ray from B. Let AB consists of n secondary sources then the phase difference between any two consecutive source will be

 

2π/λ  a sinθ = δ (say)

 

The resultant amplitude and phase at P will be

 

R = a sin nδ/2 /sin δ/2

=  a sin π a sin θ / λ / sin π a sin θ /nλ

= a sin α / sin α/n

= na  sinα/α = A sin α / α

 

A = na   and    α = πasinθ / λ

 

 

 

Corresponding the intensity is

I = R2 = A2 sin2 α / α2

Condition of maxima and minima

dI/ d α = 0 =>   d/d α [A2 sin2 α / α2] = 0

where

 

(i)   sin α / α = 0                     or                                 (ii)  α = tan α

 

Condition of minimum intensity

 

Intensity will be 0 when sin α / α = 0 or sin α = 0

 

α  = mπ

 

π a sin θ/λ = mπ

a sin θ = mλ

 

Condition of maximum intensity: Intensity will be maximum

 

a= tan a

 

The value of a satisfying this equation are obtained graphically by plotting the curve y = α and y = tan α on the same graph (Figure 5.7). The point of intersection will give

 

α = 0, ± 3π/2, ± 5 π /2, ± 7 π /2,

=0, ± 1.43 π, ± 2.462 π, ± 3.471 π,

FIGURE 5.7 Graphical representation of positions of secondary maxima’s in the diffraction

 

α = 0 correspond point to central maximum whose intensity is given as

 

I= Lt   A2 [ sin2 α / α2 = A2  = I0

 

The other maxima are given by

 

A sin θ = (2m + 1) λ/ 2

 

and their intensities as

 

m = 1:     I1 = A 2 (sin 3π/2) 2=  4 I0/ 9π2 =  I0 /22

 

m = 2:      I2 = 4I0/25π2 = I0/61

 

m3 = I3 = 4I0/ 49π2 = 10/121 and so on

 

The diffraction pattern consists of a bright central maximum surrounded alternatively by minima maximum.