In the diffraction of a single slit experiment, how would the width of and the intensity of central maximum change,if
slit width is halved and

Explanation

Consider a monochromatic source of light S, which emits light waves of wavelength `lamda`. Source S is placed at the principal focus of convex lens `L_(1)`. A paralle beam of light (a plane wave front) gets incident on a narrow slit AB of width .a.. The diffraction pattern is obtained on the screen. D= Distance between the screen and slit AB. Screen also lies at the focal plane of convex lens `L_(2)`. The diffraction pattern is found to have central maxima at the centre O of the screen, which is flanked on both sides by a number of dark and bright bands called secondary maxima and minima. Central maximum According to Huygens principle, when light falls on the slit, it becomes a source of secondary wavelets. These wavelets are initially in phase and spread out in all directions. Let C be the centre of the slit AB. The wavelets which fall on the lens in a direction parallel to CO meet at point O in the same phase. This is because, the wavelets are initially in phase and their optical paths between the slit and the point O are also equal. Therefore, the wavelets reinforce each other and give rise to the central maximum at point O. Positions and width 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 different parts of the slit do not reach point P in phase, although they are initially in phase. It is because, they cover unequal distances in reaching the point P. The wavelets from points A and B will have a path difference equal to BN. From right angled `DeltaANB`, we have
`BN =AB sin theta`
`rArr BN = a sin theta` [AB=a =Width of the slit]
Let point P is at a such a distance from centre of screen that `BN= lamda` and angle `theta= theta_(1)`
`rArr lamda= a sin theta_(1)`
`rArr sin theta_(1)= (lamda)/(a)`
It gives the positive of first secondary minima. If point P is at such a distance from the centre of the screen then `BN 2lamda and theta= theta_(2)`
then `2lamda= a sin theta_(1)`
`rArr sin theta_(2)= (2lamda)/(a)`
It gives the positive of second secondary minimum. In general for nth minimum at P, we have `sin theta_(n)= (n lamda)/(a)`
If `y_(n)`= Distance of nth minimum from the centre of screen then `tan theta_(n)= (OP)/(CO)= (y_(n))/(D)`
If `theta_(n)` is very small then `sin theta_(n)= tan theta_(n)`
`rArr sin theta_(n)= (y_(n))/(D)`
But `sin theta_(n)= (n lamda)/(a)`
`rArr y_(n)= (n D lamda)/(a)` ...(A)
The width of secondary maximum `=beta =y_(n)-y_(n-1)`
`beta= (n D lamda)/(a)- ((n-1)D lamda)/(a)`
`beta= (D lamda)/(a)` ..(*)
On the other hand, if point P on the screen is at a distance from the point O that `BN= (3 lamda)/(2)` and angle `theta= theta._(1)`, then from equation (1), we have
`(3 lamda)/(2)= a sin theta._(1)`
`sin theta._(1)= (3 lamda)/(2a)`
Such a point on the screen will be the position of first secondary maximum. In this case, the slit may be assumed to be divided into three equal parts, the wavelets from the first two parts will reach point P in opposite phase, cancelling the effect of each other. The wavelets from the third part of the slit remain uncancelled and produce the first maximum at point P. Similarly, corresponding to `BN= (5 lamda)/(2)` and angle `theta= theta._(2)`, the second secondary maximum is produced, such that `sin theta._(2)= (5lamda)/(2a)`
In this case, the slit may be assumed to be divided into five equal parts, the wavelets from first four parts cancel the effect of each other, while the wavelets from the fifth part produce the second secondary maximum at point P. In general, for nth maximum at point P, `sin theta._(n)= ((2n+1) lamda)/(2a)` ...(2)
If `y._(n)` is the distance of nth maximum from the centre of the screen, then the angular position of the nth maximum is given by
`tan theta._(n)= (y._(n))/(D)` ...(3)
In case `theta._(n)` is small, `sin theta._(n) = tan theta._(n)`
Therefore, from equations (2) and (3), we have `y._(n)= ((2n+1) D lamda)/(2a)` ...(4)
The width of the secondary minimum, `beta.= y._(n)- y._(n-1) = ((2n+1) lamda)/(2a)- ({(2n-1)-1}lamda)/(2a)`
or `beta.= (D lamda)/(a)` ...(5)
Since `beta.` is independent of n, all the secondary minima are of the same width `beta.`. Further, from equations (*) and (5), it follows that `beta= beta.` i.e., all the secondary maxima and minima are of the same width. Width of central maximum. The central maximum extends upto distance `y_(1)` (the distance of first secondary minimum) on both sides of the centre of the screen. Therefore, width of the central maximu, `beta_(0)= 2y_(1)`
In equation (A), setting n=1, we have `y_(1)= (D lamda)/(a)`
Therefore, `beta_(0) = (2 D lamda)/(a)` ...(6)
It follows that `beta_(0)=2 beta` i.e., the central maximum is twice as any other secondary maximum or minimum. From eqn (6). it is clear that fringe width is inversely proportional to the width of slit. We know width of central maximum `=beta_(0) =(2 D lamda)/(a)`
`rArr beta_(0) prop (1)/(a("width of slit"))` ...(7)
Let I= Intensity of central maximum We know that `I prop a` ...(8)
From eqns(7) and (8), it is clear that `I prop (I)/(beta_(0))`
`rArr` Intensity of central maximum is inversely proportional to the width of the slit. Fringe width depends upon the following factors:
(i) Wavelength of light. (ii) Distance of screen from slit (D). (iii) Width of the slit (a)