Question

Light containing a mixture of two wavelengths, $\mathbf{500}$ and $\mathbf{600}\mathbf{}\mathit{n}\mathit{m}$, is incident normally on a diffraction grating. It is desired (1) that the first and second maxima for each wavelength appear at$\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{⩽}{\mathbf{30}}^{\mathbf{°}}$ , (2) that the dispersion be as high as possible, and (3) that the third order for the $\mathbf{600}\mathbf{}\mathit{n}\mathit{m}$ light be a missing order.

(a) What should be the slit separation?

(b) What is the smallest individual slit width that can be used?

(c) For the values calculated in (a) and (b) and the light of wavelength$\mathbf{600}\mathbf{}\mathit{n}\mathit{m}$ , what is the largest order of maxima produced by the grating?

Short answer

(a) The slit separation such that the first and second maxima for each wavelength appear at$\sin \theta ⩽{30}^{°}$ and that the dispersion be as high as possible, is $2.4\times {10}^{-6}m$.

(b) The slit width such that the third order for the $600nm$ light is a missing order, is $8\times {10}^{-7}m$.

(c) The largest order of maxima produced by the grating is $4$.

Step-by-step solution

Given data

The wavelengths of incident light are

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

Diffraction from a grating and interference

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

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

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

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

$\mathit{a}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathit{\lambda }$ .....(2)

Step 3:Determining the slit separation

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

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

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

Determining the slit width

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

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

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

Determining the largest diffraction order

For the largest diffraction order $\sin \theta =1$. Hence from equation (I)

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

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