Find the intensities of the first three secondary maxima in the single-slit diffraction pattern, measured relative to the intensity of the central maximum

Explanation

KEY IDEAS
The secondary maxima lie approximately halfway between the minima, whose angular locations are given by Eq. 35-60 (`alpha= m pi`). The locations of the secondary maxima are then given (approximately) by
`a=(m+(1)/(2))pi`, for m=1, 2, 3, ....
with `alpha` in radian measure. We can relate the intensity I at any point in the diffraction pattern to the intensity `I_(m)` of the central maximum via Eq. 35-58.
Calculations: Substituting the approximate values of a for the secondary maxima into Eq. 35-58 to obtain the relative intensities at those maxima, we get
`(I)/(I_(m))=((sin alpha)/(alpha))^(2)=((sin(m+(1)/(2))pi)/((m+(1)/(2))pi))^(2)`, for m = 1, 2, 3,....
The first of the secondary maxima occurs for m = 1, and its relative intensity is
`(I_(1))/(I_(m))=((sin(1+(1)/(2))pi)/((1+(1)/(2))pi))^(2)=((sin 1.5pi)/(1.5pi))^(2)`
`=4.50 xx 10^(-2) ~~ 4.5%`. (Answer)
For m = 2 and m = 3, we find that
`(I_(2))/(I_(m))=1.6%` and `(I_(2))/(I_(m)) =0.83%`. (Answer)
As you can see from these results, successive secondary maxima decrease rapidly in intensity. Figure was deliberately overexposed to reveal them.