Q. Derive an expression for the intensity distribution due to Fraunhofer's diffraction at a single slit and shows that the intensity of the first subsidiary maxima is about 4.5 percent of that of the principal maxima.
Quantitative Description: Let S1 S2 is a split of width 'a'. The monochromatic light of wavelength l incident normally on this slit of width 'a'. According to Huygens's principle, each point in slit S1 S2 sends out secondary wavelets in all directions. It is clear that all the secondary starts from different points in slit S1 S2 are in the same plane. The secondary wavelets diffracted in the direction of incident ray are focused at a point Po on screen while the secondary wavelets diffracted through an angle '?' are focused at a different point P on screen. Since point Po is equidistant for all the secondary wavelets starting from slit S1 S2. So the wavelets reaching at point Po will be in phase and a maximum intensity will be found at point Po. Our purpose is to fig the resultant intensity at a given point P on screen due to secondary wavelets those are diffracted by an angle '?'.
The path difference between the secondary wavelets focusing at point P and diffracting through an angle '?' by the slit S1 S2 increases continuously from extreme points of slit S1 to S2. To evaluate this path difference between secondary wavelets emitting from S1 and S2. Let S1 K be a difference between wavelets from S1 to S2 in a direction ? can be given as
From right angle triangle S1 K S2
S2K = S1S2 sin?
S2K = a sin? ............ (I)
Equation (ii) gives the phase difference between extremum secondary wavelets starting from S1 and S2 which increases continuously from S1 to S2.
Similarly amplitude of each secondary wavelet is assumed to be a0 (let) , since each part has the same width.
Now the resultant disturbance from these entire elements can be calculated by using vector addition of polygon method or pharos diagram method. For this purpose, construct a polygon of each of equal length 'a0' and having a successive phase d/p start from point S1 and ends at S2 as shown in fig. (b).
This line S1 S2 will represent the resultant disturbance. Since the number of secondary wavelets is very large then this stepped curve can be consider as a continuous arc. The phase difference between the wavelets from extremum ends of the slit S1 S2 is the angle between tangent at point S1 & S2.
Here the resultant Amplitude S1 S2 = 2 S1 Q.
Draw a perpendicular OQ from the centre of the arc S1 S2 on chord which divides the line S1 S2 in two equal parts.
Equation (xiv) gives the expression for resultant intensity due to diffraction at a single slit.