In the two-slit interference experiment of Fig. 35-10, the slit widths are each $\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathit{\mu }\mathit{m}$, their separation is $\mathbf{24}\mathbf{.}\mathbf{0}\mathbf{}\mathit{\mu }\mathit{m}$, the wavelength is 600 nm, and the viewing screen is at a distance of 4.00 m. Let ${\mathit{I}}_{\mathbf{P}}$represent the intensity at point P on the screen, at height $\mathit{y}\mathbf{=}\mathbf{70}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{cm}}$. (a) What is the ratio of${\mathit{I}}_{\mathbf{P}}$ to the intensity${\mathit{I}}_{\mathbf{m}}$ at the centre of the pattern? (b) Determine where P is in the two-slit interference pattern by giving the maximum or minimum on which it lies or the maximum and minimum between which it lies. (c) In the same way, for the diffraction that occurs, determine where point P is in the diffraction pattern.

The expression of intensity in double slit diffraction is given by,

${\mathit{I}}_{\mathbf{P}}\mathbf{=}{\mathit{I}}_{\mathbf{m}}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\beta }\left(\frac{sin\alpha }{\alpha }\right)$ …… (1)

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

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

The expression of$\mathit{\alpha }$ is given by,

$\mathit{\alpha }\mathbf{=}\frac{\mathbf{\pi }\mathbf{a}}{\mathbf{\lambda }}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$

(a)

Calculate the value of$\theta$ as follows.

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{y}{D}\right) \\ ={\tan }^{-1}\left(\frac{0.70\text{m}}{4.00\text{m}}\right) \\ =9.926° \end{gathered}$$

Find the value of$\beta$ as follows.

role="math" localid="1663154227191" $$\begin{gathered} \beta =\frac{\pi \left(24.0\mu m\right)}{600nm}\sin \left(9.926°\right) \\ =\frac{\pi \left(24.0\mu m\right)}{600nm}\left(\frac{1nm}{{10}^{-9}m}\right)\left(\frac{{10}^{-6}m}{1\mu m}\right)\sin \left(9.926°\right) \\ =21.66rad \end{gathered}$$

Find the value of $\alpha$as follows.

$$\begin{gathered} \beta =\frac{\pi \left(12.0\mu m\right)}{600nm}\sin \left(9.926°\right) \\ =\frac{\pi \left(12.0\mu m\right)}{600nm}\left(\frac{1nm}{{10}^{-9}m}\right)\left(\frac{{10}^{-6}m}{1\mu m}\right)\sin \left(9.926°\right) \\ =10.83rad \end{gathered}$$

Substitute 21.66 rad for $\beta$, and 10.83 rad for$\alpha$ in equation (1).

$$\begin{gathered} I_P=I_m{\cos }^2\left(21.66\text{rad}\right)\left(\frac{\sin \left(10.83\text{rad}\right)}{10.83\text{rad}}\right) \\ \frac{I_P}{I_m}=7.43\times {10}^{-3} \end{gathered}$$

Therefore, the ratio is $7.43\times {10}^{-3}$.

(b)

The condition for the constructive interference maxima is given by,

$$\begin{gathered} d\sin \theta =m\lambda \\ m=\frac{d\sin \theta }{\lambda } \end{gathered}$$ ……. (2)

Substitute$24\times {10}^{-6}\text{m}$ for $d$,$9.926°$ for $\theta$, and$600\times {10}^{-9}\text{m}$ for$\lambda$ in equation (2).

$$\begin{gathered} m=\frac{\left(24\times {10}^{-6}\text{m}\right)\sin \left(9.926°\right)}{600\times {10}^{-9}\text{m}} \\ \approx 6.9 \end{gathered}$$

Therefore, the point should lie between the 6^th minima and 7^th maxima.

Thus, the point$P$ lies between the$m=6$ minimum (the seventh one), and$m=7$ maximum (the seventh side maximum).

(c)

The condition for the minima dark fringes is given by,

$$\begin{gathered} a\sin \theta =m\lambda \\ m=\frac{a\sin \theta }{\lambda } \end{gathered}$$ ……. (3)

Substitute$12\times {10}^{-6}\text{m}$ for $a$,$9.926°$ for $\theta$, and$600\times {10}^{-9}\text{m}$ for $\lambda$in equation (2).

$$\begin{gathered} m=\frac{\left(12\times {10}^{-6}\text{m}\right)\sin \left(9.926°\right)}{600\times {10}^{-9}\text{m}} \\ =3.447 \\ \approx 3.4 \end{gathered}$$

Thus, the point lies between the$m=3$ minimum (the third one), and$m=4$ maximum (the fourth one).