Question

Obtain the conditions for the bright and dark fringes in diffraction pattern due to a single narrow slit illuminated by a monochromatic source.

Explain clearly why the secondary maxima go on becoming weaker with increasing.

Answer 1

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 …(1)

Suppose BN = λ and θ = θ₁

Then, the above equation gives

λ = a sin θ₁

`sin theta_1 = lambda/a  ........ (2)`

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 theta_2 = (2lambda)/a  .......... (3)`

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  ........ (4)`

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

`tantheta_n =(OP)/(CO)=y_n/D   .............. (5)`

In case θ_n is small, sin θ_n ≈ tan θ_n

∴ Equations (iv) and (v) give

`(y_n)/D  = (nlambda)/a`

`y_n = (nlambdaD)/a`

If `BN = (3lambda)/2`and θ = θ₁′, then 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 θ = θ ′₂, the second secondary maximum is produced

In general, for the nth maximum at point P,

`sintheta_n^' = ((2n+1)lambda)/(2a)  .............. (6)`

If 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)/a  ....... (7)`

In case θ′_n is small,

sin θ’_n ≈ tan θ’_n

`y_n^' = ((2n+1)lambdaD)/(2a)`

For n = 1,

 `theta' = (3lambda)/(2a)`(From eq(6), small angle approximation, `sin theta' =theta' = ((2n+1)lambda)/(2a)` 

This angle is midway between two of the dark fringes. Divide the slit into three equal parts. If we take the first two third part of the slit the path difference between the two ends would be,

`2/3 a xx theta' = lambda`

The first two third is divided into two halves which have path difference `lambda/2`. The contribution due to these two halves is 180° out of phase and gets cancel. Only the remaining one third part of the slit contributes to the intensity at a point between the two minima which will be much weaker than the central maxima. Thus with increasing n the intensity of maxima gets weaker.