The distance between the first and fifth minima of a single slit diffraction pattern is $\mathbf{0}\mathbf{.}\mathbf{35}\mathbf{}\mathbf{mm}$with the screen $\mathbf{40}\mathbf{}\mathbf{cm}$away from the slit, when light of wavelength role="math" localid="1663070418419" $\mathbf{550}\mathbf{}\mathbf{nm}$is used. (a) Find the slit width. (b) Calculate the angle role="math" localid="1663070538179" $\mathit{\theta }$ of the first diffraction minimum.

The difference between the first and fifth minima of a single slit is $0.35\text{mm}\left(\frac{{10}^{-3}\mathrm{m}}{1\mathrm{mm}}\right)=3.5\times {10}^{-4}\mathrm{m}$.

The screen distance from slit,$\text{D}=40\text{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.40\mathrm{m}$.

The wavelength,$\lambda =550\text{nm}\left(\frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}\right)=5.50\times {10}^{-7}\mathrm{m}$

The expression for the minima in the diffraction pattern is given as follows.

$\mathbf{\text{asin}}\mathit{\theta }\mathbf{=}\mathbf{\text{m}}\mathit{\lambda }$ (1)

Here, $\mathit{\lambda }$is the wavelength, $\mathbf{m}$is the number of fringes in the envelope, androle="math" localid="1663072192166" $\mathit{\theta }$ is the angle of diffraction.

Consider the following figure on the bases of the given data

Here, $y$is the distance to minima from central maxima.

The tangential angle is given by,

$\tan \theta =\frac{y}{D}$

If the angle is very small then $\sin \theta \approx \theta \approx \tan \theta$, therefore,

$\sin \theta =\frac{y}{D}$ (2)

Recall the equation (1)

$$\begin{gathered} a\sin \theta =m\lambda \\ \sin \theta =\frac{m\lambda }{a} \end{gathered}$$

Substitute for into equation (2),

$$\begin{gathered} \frac{m\lambda }{a}=\frac{y}{D} \\ y=\frac{m\lambda D}{a} \end{gathered}$$ (3)

Calculate the distance for first order minima,

Substitute $1$for $m$into equation (3).

$$\begin{gathered} y_1=\frac{\left(1\right)\lambda D}{a} \\ {y}_1=\frac{\lambda D}{a} \end{gathered}$$

Calculate the distance for fifth order minima,

Substitute $\frac{m\lambda }{a}$for into role="math" localid="1663084231482" $\sin \theta$equation (3).

$$\begin{gathered} y_2=\frac{\left(5\right)\lambda D}{a} \\ {y}_2=\frac{5\lambda D}{a} \end{gathered}$$

Calculate the difference between the distance of first and fifth-order minima.

$$\begin{gathered} y_2-y_1=\frac{5\lambda D}{a}-\frac{\lambda D}{a} \\ {y}_2-y_1=\frac{4\lambda D}{a} \\ a=\frac{4\lambda D}{y_2-y_1} \end{gathered}$$

Substitute $0.35\text{mm}$for $y_2-y_1$, $40\text{cm}$for $D$, and $550\text{nm}$for $\lambda$into the above equation.

role="math" localid="1663085688444" $$\begin{gathered} a=\frac{4\times 5.50\times {10}^{-7}\text{m}\times 0.40\text{m}}{3.5\times {10}^{-4}\mathrm{m}} \\ a=\frac{8.8\times {10}^{-7}{\text{m}}^2}{3.5\times {10}^{-4}\mathrm{m}} \\ a=\frac{8.8\times {10}^{-3}{\text{m}}^2}{3.5\mathrm{m}} \\ a=2.51\times {10}^{-3}\text{mm}\left(\frac{1\mathrm{m}}{{10}^{-3}\mathrm{mm}}\right) \\ a=2.51\mathrm{m} \end{gathered}$$

Hence the slit width is $2.51\text{mm}$.

The angle of diffraction is given by,

$\sin \theta =\frac{m\lambda }{a}$

Substitute $1$for $\mathrm{m}$, $550\text{nm}$for $\lambda$, and $2.5\text{mm}$for a into the above equation.

role="math" localid="1663086611842" $$\begin{gathered} \sin \theta =\left(\frac{1\times 5.50\times {10}^{-7}\mathrm{m}}{2.5\times {10}^{-3}\mathrm{m}}\right) \\ \theta ={\sin }^{-1}\left(2.2\times {10}^{-4}\right) \\ \theta =0.0126° \end{gathered}$$

Convert the angle from degree to radian as

$$\begin{gathered} \theta =0.0126\times \frac{\pi }{180} \\ \theta =2.2\times {10}^{-4}\text{rad} \end{gathered}$$

Hence the angle of the first diffraction minima is $2.2\times {10}^{-4}\text{rad}$.