Show that the width of central maximum is twice that of a secondary maximum and if the width of slit is increased, the width of diffraction fringes gets diminished.

Explanation

Distance of n th minimum from central maximum
`x_(n) = (nD lambda)/(a)`
[where D = distance of slit from the screen , `lambda` = wavelength of the light, a = width of the slit]
In a diffraction pattern, secondary maxima and minima come alternatively and so the width of a secondary maximum,
`beta = x_(n) - x_(n-1) = (nDlambda)/(a) - ((n-1)Dlambda)/(a) = (Dlambda)/(a)`
Now, angular width of the central maximum,
`2theta= (2lambda)/(a)` [ see equation (6) in article 7.4]
Therefore, linear width of central maximum,
`beta_(0) = D cdot 2theta = (2Dlambda)/(a) = 2beta`
i.e,. width of central maximum is twice that of any secondary maximum.
Again, both the width of secondary maximum and central
maximum `prop (1)/("width of slit(a)").`
So, if the width of slit is increased, the width of diffraction fringes gets diminished.