Secondary maximum in Fraunhofer/single-slit diffraction

Question

Question:

The angular position at which first minima is observed is $\theta=\pi/6$ rad for a slit of width $e$ and light of wavelength $\lambda$. The angular position at which secondary maximum is observed is = ?

My attempt:

For first minima, $e\sin\theta=\lambda...(i)$ Unfortunately, formula for maxima is not given in my book so I derived it on my own:

For deriving second maxima
Consider $2N$ wavelets all throughout the slit.
Path difference between the $i$-th wavelet and $(N+i)$-th wavelet ($i\in [1,N]$) should be $\lambda$ for constructive interference.
So, if maxima occurs at $\theta'$ angular position, $\frac{e}{2}\sin\theta'=\lambda$ or $e\sin\theta'=2\lambda...(ii)$

Dividing (i) by (ii), we get:

$$\frac{\sin\theta}{\sin\theta'}=\frac{1}{2}$$ $$\Rightarrow \sin\theta'=2\sin\theta=1$$ $$\Rightarrow \theta'=\pi/2$$

which is the wrong answer -_-

My question:

Please explain my mistake and suggest correct approach.

Answer 1

Sorry for writing another answer. It appears to be to long for a comment.

Eq. $(ii)$ you wrote (in my notation) is $$\frac{u}{\pi} = \frac{e \sin \theta'}{\lambda} = 2.0$$ which turns out to be a minimum. $i$-th and $(N+i)$-th interfere constructively but then consider what happens when you superpose all the (already paired) $1,2,3, \cdots, N $ waves. The $N$-th wave is $\lambda$ behind $1$st. Therefore you have a resultant wave proportional to $$ \sum_{j=0}^{N-1} e^{2\pi i \cdot \frac{j}{N-1}}$$ which is a geometric series and evaluates to 0. Pairing many wavelets is useful for determining minima. There might be a another savvy way to determine maxima angles but I'm not aware of it. Edit: Here's the illustration using rotating vectors. Each vector has a length proportional to the amplitude of the wavelet and phase as a polar angle.

Answer 2

The full intensity (as a function of $\theta$) is given by $$ I(\theta) = \left( \frac{\sin{u}}{u} \right)^2 $$ where $ u = \frac{\pi}{\lambda}e \sin{\theta}$. You can work this out using Fraunhofer diffraction or the method of rotating vectors. Notice that the angle $\theta = \frac{\pi}{6}$ isn't 'small'. Hint: function $x \mapsto \frac{\sin x}{x}$ has maxima for $\frac{x}{\pi} = \pm 1.4303, \pm 2.4590$ etc.