(a) Describe briefly how a diffraction pattern is obtained on a screen due to a single narrow slit illuminated by a monochromatic source of light. Hence obtain the condition for the angular width of secondary maxima and secondary minima.
(b) Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture `2xx10^(-6)`m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of first maxima of the diffraction pattern obtained in the two cases.

Explanation

(a) The phenomenon of bending of light round the sharp corners of an obstacle spreading into the regions of geometrical shadow is called diffraction.


Expression For Fringe Width : Consider a parallel beam of light from a lens failing on a slit AB. As diffraction occurs, the pattern is focused on the screen XY with the help of lens L2. We will obtain a diffraction pattern that is a central maximum at the centre O flanked by a number of dark and bright fringes called secondary maxima and minima.
Central Maximum : '' Each point on the plane wave front AB sends out secondary wavelets in all directions. The waves from points equidistant from the centre C lying on the upper and lower half reach point O with zero path diffference and hence, reinforce each other, producing maximum intensity at point O.
Positions and Widths of Secondary Maxima and Minima : Consider a point P on the screen at which wavelets travelling in a direction making angle theta with CO are brought to focus by the lens. The wavelets from points A and B will have a path difference equal to BN.
From the right angled `Delta ANB`, we have:
`BN=ABsintheta`
`BN=a sintheta" "...(i)`
Suppose `BN=lambda and theta=theta_(1)`
Then, the above equation gives
`lambda = a sintheta_(1)`
`sin theta_(1)=(lambda)/(a)" "...(ii)`
Such a point on the screen will be the position of first secondary minimum.
`BN=2lambdaand theta=theta_(2)`
`2lambda=asintheta_(2)`
If `sintheta_(2)=(2lambda)/(a)" "...(iii)`
Such a point on the screen will be the position of second secondary minimum.
In general, for nth minimum at point P
`sintheta_(n)=(nlambda)/(a)" "...(iv)`
If yn is the distance of the nth minimum from the centre of the screen, from right angled ACOP, we have:
`tantheta_(n)=(OP)/(CO)`
`tantheta_(n)=(y_(n))/(D)" "...(v)`
In case `theta_(n)` is small, `sintheta_(n)=tantheta_(n)4` Equations (iv) and (v) give
`(y_(n))/(D)=(nlambda)/(a)`
`y_(n)=(Dlambda)/(a)`
Width of the secondary maximum,
`beta=y_(n)-y_(n-1)=(nDlambda)/(a)-((n-1)Dlambda)/(a)`
`beta=(Dlambda)/(a)" "...(vi)`
`therefore beta` is independent of n all the secondary maxima are of the same width `beta`.
If `BN=(3lambda)/(2)andtheta=theta'_(1)` from equation (i), we have :
`sintheta'_(1)=(3lambda)/(2a)`
Such a point on the screen will be the position of the first secondary maximum.
Corresponding to path difference,
`BN =(5lambda)/(2)and theta=theta'_(r)`
The second secondary maximum is produced. In general, for the nth maximum at point P,
`sintheta'_(n)=((2n+1)lambda)/(2a)" "...(vii)`
If `y_(n)` is the nth maximum from the center of the screen, then in angular position of the nth maximum is given by
`tantheta'_(n)=(y'_(n))/(D)" "...(viii)`
In case `theta'_(n)` is small `sintheta'_(n)~~tantheta'_(n)`
`therefore y'_(n)=((2n+1)Dlambda)/(2a)`
Width of the secondary minimum,
`beta=y'_(n)-y'_(n-1)`
`=nDlambdaa-(n-1)Dlambdaa`
`beta=(Dlambda)/(a)" "...(ix)`
Since `beta` independent of n, all the secondary minima are of the same width.
(b) For first maxima of the diffraction pattern we know :
`sintheta=3lambda_(2)a,` where is aperture of slit.
For small value of `theta, sintheta~~ tantheta = yD` where y is the distance of first minima from central line and D is the distance between slit and the screen.
So, `y=3lambda_(2)aD`
For 590 nm,
`y_(1)=3xx590xx10^(-92)xx10^(-6)xx1.5 y_(1)`
For 596nm,
`y_(2)=3xx596xx10^(-92)xx10^(-6)xx1.5 y_(2)`
`=0.6705 m`
Separate between the position of first maxima `=y_(2)=y_(1)=0.00675m`.