91Ó°ÊÓ

The wavelengths of incident light are

$$\begin{gathered} {\mathrm{λ}}_1=500nm \\ {\mathrm{λ}}_2=600nm \end{gathered}$$

The angular distance $\mathit{θ}$ of the ${\mathit{m}}_{\mathbf{t}\mathbf{h}}$ order diffraction pattern produced from a grating having line separation $\mathit{d}$ is

$\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{θ}\mathbf{=}\mathit{m}\mathit{λ}$ .....(1)

Here, $\mathit{λ}$is the wavelength of the incident light.

The angular distance $\mathit{θ}$ of the the first order interference minima for slit width $\mathit{a}$ is

$\mathit{a}\mathit{s}\mathit{i}\mathit{n}\mathit{θ}\mathbf{=}\mathit{λ}$ .....(2)

For maximum dispersion the second maxima of ${\mathrm{λ}}_2$should be at ${30}^{°}$. Thus from equation (1)

$$\begin{gathered} d\sin 30°=2×600nm \\ d=\frac{2×600nm}{\sin {30}^{°}} \\ =2400nm \\ =2.4×{10}^{-6}m \end{gathered}$$

Thus, the slit separation is$2.4×{10}^{-6}m$ .

The third order for ${\mathrm{λ}}_2$ is a missing order. This is because it coincides with the first interference minima. Thus from equations (1) and (2)

$$\begin{gathered} \frac{3{\mathrm{λ}}_2}{d}=\frac{{\mathrm{λ}}_2}{a} \\ a=\frac{d}{3} \\ =\frac{2.4×{10}^{-6}m}{3} \\ =8×{10}^{-7}m \end{gathered}$$

The slit width is$8×{10}^{-7}m$$8×{10}^{-7}m$.

For the largest diffraction order $\sin \mathrm{θ}=1$. Hence from equation (I)

$$\begin{gathered} d=m_{max}{\mathrm{λ}}_2 \\ {m}_{max}=\frac{d}{\mathrm{λ}} \\ =\frac{2.4×{10}^{-6}m}{600×{10}^{-9}m} \\ =4 \end{gathered}$$

Thus, the maximum order of diffraction is$4$ .