91Ó°ÊÓ

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

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,$\mathrm{λ}=550\text{nm}\left(\frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}\right)=5.50×{10}^{-7}\mathrm{m}$

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

$\mathbf{\text{asin}}\mathit{θ}\mathbf{=}\mathbf{\text{m}}\mathit{λ}$ (1)

Here, $\mathit{λ}$is the wavelength, $\mathbf{m}$is the number of fringes in the envelope, androle="math" localid="1663072192166" $\mathit{θ}$ 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,

$\tan \mathrm{θ}=\frac{y}{D}$

If the angle is very small then $\sin \mathrm{θ}≈\mathrm{θ}≈\tan \mathrm{θ}$, therefore,

$\sin \mathrm{θ}=\frac{y}{D}$ (2)

Recall the equation (1)

$$\begin{gathered} a\sin \mathrm{θ}=m\mathrm{λ} \\ \sin \mathrm{θ}=\frac{m\mathrm{λ}}{a} \end{gathered}$$

Substitute for into equation (2),

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

Calculate the distance for first order minima,

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

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

Calculate the distance for fifth order minima,

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

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

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

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

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

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

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

The angle of diffraction is given by,

$\sin \mathrm{θ}=\frac{m\mathrm{λ}}{a}$

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

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

Convert the angle from degree to radian as

$$\begin{gathered} \mathrm{θ}=0.0126×\frac{\mathrm{π}}{180} \\ \mathrm{θ}=2.2×{10}^{-4}\text{rad} \end{gathered}$$

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