How to calculate the width of the dark fringes? I am talking about the single slit diffraction experiment.
The width of the central bright fringe is twice as wide as that of the others bright fringes. It can be calculated easily as follows.
\begin{align}
\text{width of other bright fringes} = \frac{\text{wave length}\times\text{distance between screen and slit}}{\text{width of the slit}}
\end{align}

Question
From the intensity plot, it is clear that the width of dark fringes is zero. But when we look at the spectrum the width is not exactly zero. Could you tell me how to find the width of the dark fringes? I think several percentage of the maximum intensity should be considered as dark, right? What percentage is it usually adopted by physicists?
 A: This is actually somewhat more involved. You can compute the intensity/irradiance of the Fraunhofer diffraction pattern exactly and the result is the so-called Airy pattern:
$I(\theta)=I_0\cdot \left[ \frac{2 \cdot J_1(k \cdot a \cdot sin \theta)}{k \cdot a \cdot sin \theta} \right]^2$,
where $\theta$ is the observation angle, $k$ is the wavenumber and $a$ is the size of your aperture. This gives you the following intensity graph you have shown in your question.
I am referencing the Wikipedia article here. In order to evaluate it you need to be able to compute Bessel functions of the first kind and these are available in most programming languages, such as Python and Fortran, as well as software like Matlab.
The angles where the intensity minima occur are the zeros of these Bessel functions $J_1(x)$. Starting from there you can e.g. compute the angle $\theta$ at which the first intensity minimum occurs:
$sin(\theta) \approx \frac{\lambda}{d}$,
where $\lambda$ is your wavelength and $d$ the width of your aperture.
More zeros can be found on this Mathematica website. Here is the list of the first five zeros of the Bessel function of the first kind for reference:


*

*3.8317

*7.0156

*10.1735

*13.3237

*16.4706

*...


So e.g. for the first intensity minimum you would have to solve $k \cdot a \cdot sin(\theta) = 3.8317$ for $\theta$, which is:
$\theta = arcsin \left( \frac{3.8317}{k\cdot a} \right)$
This also means that the intensity is zero at specific angles and not over a continuum. It only appears so due to the contrast of the image you have chosen.
