Question

If a diffraction grating produces its third-order bright band at an angle of $\mathbf{78}\mathbf{.}\mathbf{4}\mathbf{°}$ for light of wavelength 681nm, find (a) the number of slits per centimeter for the grating and (b) the angular location of the first-order and second-order bright bands. (c) Will there be a fourth-order bright band? Explain.

Short answer

1. There are 4790 slits per centimeter for the grating.
2. The angular location of the first-order and second-order bright bands is$19.1°\mathrm{and}40.8°$ respectively.
3. There won’t be any fourth-order bright band.

Step-by-step solution

Formulas used to solve the question

Number of slits per meter

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

Angular position of a bright band

$\mathit{s}\mathit{i}\mathit{n}{\mathit{\theta }}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{m\lambda }}{\mathbf{d}}$ (2)

Determine the number of slits per centimeter

Given: $$\begin{gathered} {\theta }_3=78.4^{\circ } \\ \lambda =681nm=681*{10}^{-9}m \end{gathered}$$

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

$d=\frac{3\lambda }{sin78.4^{\circ }}=\frac{3*681}{0.98}=2085*{10}^{-4}m$

Now, the number of slits is given as

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

Determine the angular location of the first-order and second-order bright bands

From equation (2),

For first-order bright band,

$$\begin{gathered} \sin {\theta }_1=\frac{\lambda }{d}=0.33 \\ \Rightarrow {\theta }_1={\sin }^{-1}0.33=19.1^{\circ } \end{gathered}$$

For second-order bright band,

$$\begin{gathered} sin{\theta }_2=\frac{2\lambda }{d}=0.66 \\ \Rightarrow {\theta }_2={\sin }^{-1}0.66=40.8^{\circ } \end{gathered}$$

Determine if there is any fourth-order bright band

To have $m=4\mathrm{then},\sin {\theta }_4=4\sin {\theta }_1=1.32$ 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 $19.1^{\circ }\mathrm{and}40.8^{\circ }$ respectively. There won’t be any fourth-order bright band.