Light of wavelength 440 nm passes through a double slit, yielding a diffraction pattern whose graph of intensity I versus angular position is shown in Fig. 36-44. Calculate (a) the slit width and (b) the slit separation. (c) Verify the displayed intensities of the $\mathit{m}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}$and $\mathit{m}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}$ interference fringes.

The equation for diffraction minimum is given by,

$$\begin{gathered} \mathit{a}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathit{m}\mathit{\lambda } \\ \mathit{a}\mathbf{=}\frac{\mathbf{m}\mathbf{\lambda }}{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right) \end{gathered}$$

Here,$\mathit{\lambda }$is wavelength,$\mathit{\theta }$is angle of diffraction, and a is slit width.

The expression of intensity of diffraction pattern at any angle is given by,
$\mathit{I}\left(\theta \right)\mathbf{=}{\mathit{I}}_{\mathbf{m}}\left(\frac{sin\alpha }{\alpha }\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(2\right)$

Here, ${\mathit{I}}_{\mathbf{m}}$is the maximum intensity, and $\mathit{\alpha }\mathbf{=}\frac{\mathbf{\pi }\mathbf{a}}{\mathbf{\lambda }}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$.

Substitute$5.0°$ for$\theta$ ,$440\times {10}^{-9}\text{m}$ for$\lambda$ , and 1 for$m$ in equation (1).

$$\begin{gathered} a=\frac{\left(1\right)\left(440\times {10}^{-9}\text{m}\right)}{\sin \left(5.0°\right)} \\ \approx 5.0\times {10}^{-6}\text{m} \\ =5.0\mu m \end{gathered}$$

Therefore, the slit width is$5.0\mu m$ .

The number of spots is the ratio of $d$to $a$. Here, $d$is the slit separation.

From the given figure, the total number of spots is $9$, and is given as follows.

$$\begin{gathered} \left(\frac{d}{a}\right)+1=9 \\ d=4a \\ =4\left(5.0\mu m\right) \\ =20\mu m \end{gathered}$$

Therefore, the slit separation is $20\mu m$.

From the graph, the first interference maximum is occurred at$1.25°$.

Find the value of$\alpha$as follows.

$$\begin{gathered} \alpha =\frac{\pi \left(5.0\times {10}^{-6}\text{m}\right)\left(\sin 1.25°\right)}{440\times {10}^{-9}\text{m}} \\ =0.7787\text{rad} \end{gathered}$$

Find the intensity at the second interference maximum.

$$\begin{gathered} I=\left(7mW/{\text{cm}}^{\text{2}}\right){\left(\frac{\sin \left(0.7787\text{rad}\right)}{0.7787\text{rad}}\right)}^2 \\ =5.7mW/{\text{cm}}^{\text{2}} \end{gathered}$$

Thus, the intensity at $m=1$is$5.7mW/{\text{cm}}^{\text{2}}$.

From the graph, the second interference maximum is occurred at $2.5°$.

Find the value of $\alpha$as follows.

$$\begin{gathered} \alpha =\frac{\pi \left(5.0\times {10}^{-6}\text{m}\right)\left(\sin 2.5°\right)}{440\times {10}^{-9}\text{m}} \\ =1.557\text{rad} \end{gathered}$$

Find the intensity at the second interference maximum.

$$\begin{gathered} I=\left(7mW/{\text{cm}}^{\text{2}}\right){\left(\frac{\sin \left(1.557\text{rad}\right)}{1.557\text{rad}}\right)}^2 \\ =2.9mW/{\text{cm}}^{\text{2}} \end{gathered}$$

Thus, the intensity at$m=2$ is$2.9mW/{\text{cm}}^{\text{2}}$ .

Therefore, the values of intensities of interference fringes are agreed with the values given in the graph.