diffractionopticswaves
The width of a minimum in single-slit diffraction is related to
$\ d \sin \theta=n \lambda$
where $\sin\theta = \frac{y}{L}$ by small angle approximation.
However, Wikipedia says that there is no such formula for the width of an n-th maximum in single-slit diffraction. Why is that so?
You need to split the slit by small zones and sum up field from the zones. If the first and the last zones has destructive interference with each other than we have big total intensity. If the phase make a full turn we have a dark line. Therefore, angles of the dark lines are $\theta_n = n\lambda/b,\ \,n = 1,2,3,\dots$
any angle over theta 90 will mean that the diffraction will be going behind the diffraction gratings which is impossible. so 90 is the maximum that you can get this is why you have to round down the decimal answer you will get.
Best Answer
The problem in single slit diffraction is that there is an exception when considering the maxima. The central maximum is actually twice as wide as the other maxima. If you take this exception into account however the same formula that is valid for the minima is also valid for the maxima.
All minima have a width
$\Delta y = \lambda L /d$
The same applies to the maxima except for the centre which has a width of
$\Delta y = 2 \lambda L /d$