Question

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 conditions for the angular width of secondary maxima and secondary minima.

Answer 1

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

Expression For Fringe Width

Consider a parallel beam of light from a lens falling 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 difference 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 θ 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 ΔANB, we have:

BN = AB sinθ

BN = a sinθ …(i)

Suppose BN = λ and θ = θ₁

Then, the above equation gives

λ = a sin θ₁

`sintheta_1=lambda/a`

Such a point on the screen will be the position of first secondary minimum

If BN = 2λ and θ= θ₂, then

2λ = a sin θ₂

`sin_2=(2lambda)/a `

Such a point on the screen will be the position of second secondary minimum.

In general, for n^th minimum at point P,

`sintheta_n=(nlambda)/a`

If y_n is the distance of the n^th minimum from the centre of the screen, from right-angled ΔCOP, we have:

`tantheta_n=(OP)/(CO)`

`tantheta_n=y_n/D`

In case θ_n is small, sin θ_n ≈ tan θ_n

∴ Equations (iv) and (v) give

`y_n/D=(nlambda)/a`

`y_n=(nDlambda)/a`

Width of the secondary maximum,

`beta=y_n-y_(n-1)=(nDlambda)/a-((n-1)Dlambda)/a`

`beta=(Dlambda)/a ...(vi)`

∴ β is independent of n, all the secondary maxima are of the same width β.

If `BN=(3lambda)/2`and `theta=theta_1^'`, from equation (i), we have:

`sin_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_2^'`the second secondary maximum is produced

In general, for the n^th maximum at point P,

`sin_n^'=((2n+1)lambda)/(2a) `

If `y_n^'`is the distance of n^th maximum from the centre of the screen, then the angular position of the n^th maximum is given by

`tantheta_n^'=y_n^'/D `

In case `theta_n^'`is small

`sintheta_n^' ~~tantheta_n^'`

`:.y_n^'=((2n+1)Dlambda)/(2a)`

Width of the secondary minimum,

β' = y'_n − y'_n−1 = `(nDlambda)/a-((n-1)Dlambda)/a`

 `beta'=(Dlambda)/a `

 Since β' is independent of n, all the secondary minima are of the same width β'