91Ó°ÊÓ

Number of slits per meter

$\mathit{N}\mathbf{=}\frac{\mathbf{1}}{\mathbf{d}}$ (1)

Angular position of a bright band

$\mathit{s}\mathit{i}\mathit{n}{\mathit{θ}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{³¾Î»}}{\mathbf{d}}$ (2)

Given: $$\begin{gathered} {\mathrm{θ}}_3=78.4^{∘} \\ \mathrm{λ}=681nm=681*{10}^{-9}m \end{gathered}$$

From equation (2), solve forand plug the given,

$d=\frac{3\mathrm{λ}}{sin78.4^{∘}}=\frac{3*681}{0.98}=2085*{10}^{-4}m$

Now, the number of slits is given as

$N=\frac{1}{2085*{10}^{-4}}=4790$

From equation (2),

For first-order bright band,

$$\begin{gathered} \sin {\mathrm{θ}}_1=\frac{\mathrm{λ}}{d}=0.33 \\ ⇒{\mathrm{θ}}_1={\sin }^{-1}0.33=19.1^{∘} \end{gathered}$$

For second-order bright band,

$$\begin{gathered} sin{\mathrm{θ}}_2=\frac{2\mathrm{λ}}{d}=0.66 \\ ⇒{\mathrm{θ}}_2={\sin }^{-1}0.66=40.8^{∘} \end{gathered}$$

To have $m=4\mathrm{then},\sin {\mathrm{θ}}_4=4\sin {\mathrm{θ}}_1=1.32$ which is above that is a maximum value can take. So, there won’t be any fourth-order bright band.

Thus, there are 4790 s lits per centimeter for the grating. The angular location of the first-order and second-order bright bands is $19.1^{∘}\mathrm{and}40.8^{∘}$ respectively. There won’t be any fourth-order bright band.