Wavelength used in single slit diffraction is 628nm. Width of slit is 0.2mm. Find the angular width of central maxima in degree.

To find the angular width of the central maximum in a single-slit diffraction pattern, we use the relevant wave optics formula.

In single-slit diffraction, the first minima occur at angles given by the condition:
$d\sin \left(\theta \right)=\pm \lambda$
where:
- $d$ is the width of the slit.
- $\lambda$ is the wavelength of the light used.
- $\theta$ is the angle of the first diffraction minimum.

For a small angle $\theta$, we can approximate $\sin \left(\theta \right)\approx \theta$ (in radians). Thus:
$\theta \approx \frac{\lambda }{d}$

The central maximum lies between the first minimum on one side and the first minimum on the other side. Therefore, the total angular width of the central maximum ($2\theta$) in radians is:
$2\theta =\frac{2\lambda }{d}$

Let's write down the given values:
Wavelength, $\lambda =628\text{nm}=628\times {10}^{-9}\text{m}$
Slit width, $d=0.2\text{mm}=0.2\times {10}^{-3}\text{m}=2\times {10}^{-4}\text{m}$

Now, calculate the angular width in radians:
$2\theta =\frac{2\times 628\times {10}^{-9}}{2\times {10}^{-4}}$
$2\theta =628\times {10}^{-5}\text{rad}=0.00628\text{rad}$

To convert this value from radians to degrees, we multiply by $\frac{180}{\pi }$ (where $\pi \approx 3.14$):
$2\theta =0.00628\times \frac{180}{3.14}\text{degrees}$

Since $0.00628=2\times 3.14\times {10}^{-3}$, we can simplify the expression:
$2\theta =2\times {10}^{-3}\times 180$
$2\theta =0.36°$

Thus, the angular width of the central maxima in degrees is 0.36°.