91Ó°ÊÓ

The expression for the slit width of the diffraction grating is given by,

$$\begin{gathered} \mathit{β}\mathbf{=}\frac{\mathbf{π}\mathbf{d}}{\mathbf{λ}}\mathit{s}\mathit{i}\mathit{n}\mathit{θ} \\ \mathit{d}\mathbf{=}\mathbf{\left(}\frac{\mathbf{β}}{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{θ}}\mathbf{\right)}\mathbf{\left(}\frac{\mathbf{λ}}{\mathbf{π}}\mathbf{\right)} \end{gathered}$$ …… (1)

Here, d is slit separation,$\mathit{β}$is slit width,$\mathit{λ}$is wavelength, and$\mathit{θ}$is angle of diffraction.

From the graph, the ratio of slit with to sine of angle is,

$\begin{array}{rcl}\frac{\mathrm{β}}{\sin \mathrm{θ}}& =& \frac{80\text{rad}}{1}\\ & =& 80\text{rad}\end{array}$

Substitute$80\text{rad}$for $\frac{\mathrm{β}}{\sin \mathrm{θ}}$, and$435×{10}^{-9}\text{m}$for in equation (1).

$\begin{array}{rcl}d& =& \left(80\text{rad}\right)\left(\frac{435×{10}^{-9}\text{m}}{\mathrm{π}}\right)\\ & =& 11.1×{10}^{-6}\text{m}\\ & =& 11.1\mathrm{μ}m\end{array}$

Therefore, the slit separation is$11.1\mathrm{μ}m$.

The condition for the constructive interference maxima is given by,

$\begin{array}{rcl}d\sin {\mathrm{θ}}_{\max }& =& m_{\max }\mathrm{λ}\\ m_{\max }& =& \frac{d\sin {\mathrm{θ}}_{\max }}{\mathrm{λ}}\end{array}$ …… (2)

Here,${\mathrm{θ}}_{\max }$is the maximum angle, and$m_{\max }$is the number of maxima.

Substitute $11.1×{10}^{-6}\text{m}$for$d$,$90°$for ${\mathrm{θ}}_{\max }$, and$435×{10}^{-9}\text{m}$for$\mathrm{λ}$in equation (2).

$\begin{array}{rcl}{m}_{\max }& =& \frac{\left(11.1×{10}^{-6}\text{m}\right)\sin 90°}{435×{10}^{-9}\text{m}}\\ & =& 25.52\\ & ≈& 25\end{array}$

The number of interference maxima are 25 and the central maxima is formed at 0 . So, the number of interference minima are 25 .

The total number of interference maxima and minima on both sides of the interference pattern is,

$$\begin{gathered} n=25+25 \\ =50 \end{gathered}$$

Therefore, the total number of interference maxima on both sides of the interference pattern is 50.

The least maxima is occurred at the center if the interference pattern.

$m_{\max }=0$

From equation (1),

$$\begin{gathered} {\mathrm{θ}}_{\max }={\sin }^{-1}\left(\frac{m_{\max }\mathrm{λ}}{d}\right) \\ ={\sin }^{-1}\left(\frac{\left(0\right)\mathrm{λ}}{d}\right) \\ =0° \end{gathered}$$

Therefore, the smallest angle for maxima is$0°$ .

The number of minima is 25. So, the smallest minima is occurred at 1 and the greatest minima is occurred at 25 .

$m_{\min }=25$

The condition for interference minima is given by,

$$\begin{gathered} d\sin {\mathrm{θ}}_{\min }=m_{\min }\mathrm{λ} \\ {\mathrm{θ}}_{\min }={\sin }^{-1}\left(\frac{m_{\min }\mathrm{λ}}{d}\right) \end{gathered}$$ …… (3)

Substitute$11.1×{10}^{-6}\text{m}$ for$d$ ,$25$ for$m_{\min }$ , and$435×{10}^{-9}\text{m}$ for$\mathrm{λ}$ in equation (3).

$$\begin{gathered} {\mathrm{θ}}_{\min }={\sin }^{-1}\left(\frac{\left(25\right)\left(435×{10}^{-9}\text{m}\right)}{11.1×{10}^{-6}\text{m}}\right) \\ =78.44° \end{gathered}$$

Therefore, the greatest angle for minima is$78.44°$ .