A diffraction grating having is illuminated with a light signal containing only two wavelengths and . The signal in incident perpendicularly on the grating. (a) What is the angular separation between the second order maxima of these two wavelengths? (b) What is the smallest angle at which two of the resulting maxima are superimposed? (c) What is the highest order for which maxima of both wavelengths are present in the diffraction pattern?

The given data is listed below as-

The wavelength of light, ${\lambda }_1=400\text{nm}$

The wavelength of light, ${\lambda }_2=500\text{nm}$

Distance between two adjacent rulings, $d=\frac{1\text{mm}}{180}=5.55\times {10}^{-6}\text{m}$

In case of diffraction experiment, a maxima is formed at that point on the screen, for which the path difference between the rays is an integral multiple of wavelength of light used.

$\mathbf{d}\mathbf{sin\theta }\mathbf{=}\mathbf{m\lambda }\left(\mathbf{m}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}.....\right)······················\left(1\right)$

Here, $d$is the distance between adjacent rulings, $m$is an integer and$\lambda$is the wavelength of light used.

(a)

Using equation (1), angular separation for second order maxima $\left(m=2\right)$ is given by-

$\theta ={\sin }^{-1}\left(\frac{2\lambda }{d}\right)$

Here, $\lambda$is the wavelength and $d$is the distance between two adjacent rulings on the grating.

For, ${\lambda }_1=400\text{nm}$and $d=5.55\times {10}^{-6}\text{m}$

$$\begin{gathered} {\theta }_1={\sin }^{-1}\left(\frac{2\lambda }{d}\right) \\ {\theta }_1={Sin}^{-1}\left(\frac{2\times 400\times {10}^{-9}\text{m}}{5.5\times {10}^{-6}\text{m}}\right) \\ {\theta }_1=8.36° \end{gathered}$$

Similarly, for ${\lambda }_2=500nm$and $d=\frac{1}{180}=5.55\times {10}^{-6}m$

$$\begin{gathered} {\theta }_2={\sin }^{-1}\left(\frac{2\lambda }{d}\right) \\ {\theta }_2={Sin}^{-1}\left(\frac{2\times 500\times {10}^{-9}}{5.5\times {10}^{-6}}\right) \\ {\theta }_2=10.48° \end{gathered}$$

Now, change in angle is given by-

$$\begin{gathered} \Delta \theta ={\theta }_2-{\theta }_1 \\ \Delta \theta =10.48°-8.36° \\ \Delta \theta =2.12° \end{gathered}$$

Thus, the angular separation between the second order maxima of these two wavelengths is$\Delta \theta =2.12°$ .

(b)

When the two resulting maxima are superimposed, applying equation (1) for both the wavelengths, we get-

$m_1{\lambda }_1=m_2{\lambda }_2$

Now, find the two values of satisfying the above conditions.

For, $m_1=5$

$$\begin{gathered} m_1{\lambda }_1=5\times 400\times {10}^{-9}\text{m} \\ \text{= 2}\times {10}^{-6}\text{m} \end{gathered}$$

And for, $m_2=4$

$$\begin{gathered} m_2{\lambda }_2=4\times 500\times {10}^{-9}\text{m} \\ \text{= 2}\times {10}^{-6}\text{m} \end{gathered}$$

Now, the smallest angle is given by-

$$\begin{gathered} \varphi ={\sin }^{-1}\left(\frac{m_1{\lambda }_{}}{d}\right) \\ \varphi ={Sin}^{-1}\left(\frac{5\times 400\times {10}^{-9}\text{m}}{5.55\times {10}^{-6}\text{m}}\right) \\ \varphi =21.1° \end{gathered}$$

Thus, the smallestangle at which two of the resulting maxima are superimposed is $21.1°$

(c)

The maxima of the diffraction gratingis formed when –

$d\sin \left(\theta \right)=m\lambda$

Put $\sin \theta =1$ to find the largest value of m.

So, $d=m_{\max }\lambda$

For, $d=5.55\times {10}^{-6}\text{m}$ and ${\lambda }_2=500\text{nm}$

The largest value of m is:

$$\begin{gathered} m_{\max }=\frac{d}{{\lambda }_2} \\ {m}_{\max }=\frac{5.55\times {10}^{-6}\text{m}}{500\times {10}^{-9}\text{m}} \\ {m}_{\max }=11 \end{gathered}$$

Thus, the highest order for which maxima of both wavelengths are present in the diffraction pattern is $11$.