(a) How many bright fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern if $\mathit{\lambda }\mathbf{=}\mathbf{550}\mathbf{}\mathit{n}\mathit{m}$, $\mathit{d}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{150}\mathbf{}\mathit{m}\mathit{m}$, and $\mathit{a}\mathbf{=}\mathbf{30}\mathbf{}\mathit{\mu }\mathit{m}$? (b) What is the ratio of the intensity of the third bright fringe to the intensity of the central fringe?

Wavelength,$\lambda =550\text{nm}$

The slit separation,$d=0.150\text{mm}$

The slit width,$a=30\mu \text{m}$

The expression to calculate the location of first minima is single slit diffraction is given as follows.

$\mathit{a}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathit{\lambda }$ …… (i)

The expression to calculate the location of first minima is double slit diffraction is given as follows.

$\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathit{m}\mathit{\lambda }$ …… (ii)

The expression to calculate the intensity of the bright fringes is given as follows.

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

Here,${\mathit{I}}_{\mathbf{m}}$ is the intensity at the centre of the pattern.

(a)

Recall the equation (i).

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

Recall equation (ii).

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

Equate the equation (iii) and (iv).

$$\begin{gathered} \frac{\lambda }{a}=\frac{m\lambda }{d} \\ \frac{1}{a}=\frac{m}{d} \\ m=\frac{d}{a} \end{gathered}$$

Substitute$0.150\text{mm}$ for$d$ and$30\mu \text{m}$ for$a$ into above equation.

$$\begin{gathered} m=\frac{0.150\times {10}^{-3}}{30\times {10}^{-6}} \\ m=\frac{150\times {10}^{-6}}{30\times {10}^{-6}} \\ m=5 \end{gathered}$$

The$m=5$ is the number of the fringes only in side of the centre pattern, therefore the total number of the bright fringe is given by,

$N=2\left(m-1\right)+1$

Substitute 5 for $m$into above equation.

$$\begin{gathered} N=2\left(5-1\right)+1 \\ N=2\times 4+1 \\ N=8+1 \\ N=9 \end{gathered}$$

Hence the number of the bright fringes is 9.

(b)

For third bright fringe, $m=3$

Recall equation (ii),

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

The value of the $\alpha$can be calculated as,

$\alpha =\frac{\pi a}{\lambda }\sin \theta$

Substitute$\frac{m\lambda }{d}$ for$\sin \theta$ into above equation.

$$\begin{gathered} \alpha =\frac{\pi a}{\lambda }\times \frac{m\lambda }{d} \\ \alpha =\frac{m\pi a}{d} \end{gathered}$$

Substitute 3 for$m$ ,$0.150\text{mm}$ for$d$ and$30\mu \text{m}$ for$a$ into above equation.

$$\begin{gathered} \alpha =\frac{3\times \pi \times 30\times {10}^{-6}}{0.150\times {10}^{-3}} \\ \alpha =\frac{282.743\times {10}^{-6}}{150\times {10}^{-6}} \\ \alpha =1.884\text{rad} \end{gathered}$$

The value of the$\beta$ can be calculated as,

$\beta =\frac{\pi d}{\lambda }\sin \theta$

Substitute$\frac{m\lambda }{d}$ for$\sin \theta$ into above equation.

$$\begin{gathered} \beta =\frac{\pi d}{\lambda }\times \frac{m\lambda }{d} \\ \beta =m\pi \end{gathered}$$

Substitute 3 for$m$ into above equation.

$$\begin{gathered} \beta =3\pi \\ \beta =9.424 \end{gathered}$$

Calculate the ratio of the intensity of the third and centre fringe.

$$\begin{gathered} I_3\left(\theta \right)=I_m{\cos }^2\left(9.424\right)\times {\left(\frac{\sin 1.884}{1.884}\right)}^2 \\ \frac{I_3\left(\theta \right)}{I_m}=0.9662\times 0.2661 \\ \frac{I_3\left(\theta \right)}{I_m}=0.257 \end{gathered}$$

Hence the ratio of the intensity of bright third and centre fringe is 0.257.