Obtain the formulas for the angular width and linear width of the central maximum in a single-slit diffraction experiment.

(N/A) The width of the central maximum is defined as the distance between the first minima on either side of the central maximum.
Let the width of the slit be $a$ and the distance between the slit and the screen be $D$.
For the first minimum on one side of the central maximum,the condition for diffraction is given by $a \sin \theta = \lambda$. Since $\theta$ is very small,$\sin \theta \approx \theta$,so $\theta = \frac{\lambda}{a}$.
Here,$\theta$ represents half the angular width of the central maximum.
Therefore,the total angular width of the central maximum is $2\theta = \frac{2\lambda}{a}$.
Now,for the linear width $\beta_0$,we use the relation between arc length,radius,and angle: $\text{arc} = \text{radius} \times \text{angle}$.
Here,the arc length is $\beta_0$,the radius is $D$,and the angle is $2\theta$.
Thus,$\beta_0 = D \times (2\theta) = D \times \frac{2\lambda}{a}$.
Therefore,the linear width of the central maximum is $\beta_0 = \frac{2D\lambda}{a}$.