$\mathit{F}\mathit{i}\mathit{g}\mathit{u}\mathit{r}\mathit{e}\mathbf{}\mathbf{36}\mathbf{-}\mathbf{38}$gives $\mathit{\alpha }$versus the sine of the angle $\mathit{\theta }$in a single-slit diffraction experiment using light of wavelength $\mathbf{610}\mathbf{}\mathbf{nm}$. The vertical axis scale is set by as role="math" localid="1663169810058" ${\mathbf{\alpha }}_{\mathbf{s}}\mathbf{=}\mathbf{12}\mathbf{}\mathbf{rad}$. What are (a) the slit width, (b) the total number of diffraction minima in the pattern (count them on both sides of the center of the diffraction pattern), (c) the least angle for a minimum, and (d) the greatest angle for a minimum?

The wavelength of the light is $\lambda =610\text{nm}$.

The angle on vertical axis is ${\alpha }_s=12\text{rad}$.

The angle of the diffraction is defined as the angle of central axis from the diffraction minima of different orders.

(a)

The slit width is given as:

${\alpha }_s=\frac{\pi d}{\lambda }$

Substitute all the values in equation.

$$\begin{gathered} 12\text{rad}=\frac{\pi d}{610\text{nm}} \\ d=2330.03\text{nm} \end{gathered}$$

Therefore, the slit width is $2330.03\text{nm}$.

(b)

The total number of diffraction minima in diffraction pattern is given as:

$m\lambda =d\sin \theta$

Here, $\theta$is diffraction angle and its value for total number of minima is $90°$. because beyond this angle diffraction is not observed.

Substitute all the values in equation.

$$\begin{gathered} m\left(610\text{nm}\right)=\left(2330.03\text{nm}\right)\left(\sin 90°\right) \\ m=3.82 \\ m\approx 4 \end{gathered}$$

Therefore, the total number of diffraction minima in diffraction pattern is $4$.

(c)

The least angle for minimum is given as:

$\sin {\theta }_{\min }=\frac{m\lambda }{d}$

The value of m is 1 for least angle for minimum.

Substitute all the values in equation.

$$\begin{gathered} \sin {\theta }_m=\frac{\left(1\right)\left(610\text{nm}\right)}{2330.03\text{nm}} \\ {\theta }_{\min }=15.2° \end{gathered}$$

Therefore, the least angle for minimum is $15.2°$.

(d)

The greatest angle for minimum is given as:

$\sin {\theta }_{\max }=\frac{m\lambda }{d}$

The value of m is 3 for greatest angle for minimum.

Substitute all the values in equation.

$$\begin{gathered} \sin {\theta }_{\max }=\frac{\left(3\right)\left(610\text{nm}\right)}{2330.03\text{nm}} \\ {\theta }_{\max }=51.2° \end{gathered}$$

Therefore, the greatest angle for minimum is $51.2°$.