91Ó°ÊÓ

The given data is listed below as-

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

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

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

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.

$\mathbf{d}\mathbf{²õ¾\pm ²Ôθ}\mathbf{=}\mathbf{³¾Î»}\left(\mathbf{m}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}.....\right)······················\left(1\right)$

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

(a)

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

$\mathrm{θ}={\sin }^{-1}\left(\frac{2\mathrm{λ}}{d}\right)$

Here, $\mathrm{λ}$is the wavelength and $d$is the distance between two adjacent rulings on the grating.

For, ${\mathrm{λ}}_1=400\text{nm}$and $d=5.55×{10}^{-6}\text{m}$

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

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

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

Now, change in angle is given by-

$$\begin{gathered} \mathrm{Δ}\mathrm{θ}={\mathrm{θ}}_2-{\mathrm{θ}}_1 \\ \mathrm{Δ}\mathrm{θ}=10.48°-8.36° \\ \mathrm{Δ}\mathrm{θ}=2.12° \end{gathered}$$

Thus, the angular separation between the second order maxima of these two wavelengths is$\mathrm{Δ}\mathrm{θ}=2.12°$ .

(b)

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

$m_1{\mathrm{λ}}_1=m_2{\mathrm{λ}}_2$

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

For, $m_1=5$

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

And for, $m_2=4$

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

Now, the smallest angle is given by-

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

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

(c)

The maxima of the diffraction gratingis formed when –

$d\sin \left(\mathrm{θ}\right)=m\mathrm{λ}$

Put $\sin \mathrm{θ}=1$ to find the largest value of m.

So, $d=m_{\max }\mathrm{λ}$

For, $d=5.55×{10}^{-6}\text{m}$ and ${\mathrm{λ}}_2=500\text{nm}$

The largest value of m is:

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

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