A beam of light consisting of wavelengths from$\mathbf{460}\mathbf{.}\mathbf{0}\mathbf{}\mathit{n}\mathit{m}\mathbf{}\mathit{t}\mathit{o}\mathbf{}\mathbf{640}\mathbf{.}\mathbf{0}\mathbf{}\mathit{n}\mathit{m}$is directed perpendicularly onto a diffraction grating with 160 lines/mm. (a) What is the lowest order that is overlapped by another order? (b) What is the highest order for which the complete wavelength range of the beam is present? In that highest order, at what angle does the light at wavelength (c)$\mathbf{460}\mathbf{.}\mathbf{0}\mathbf{}\mathit{n}\mathit{m}\mathbf{}$and (d) $\mathbf{640}\mathbf{.}\mathbf{0}\mathbf{}\mathit{n}\mathit{m}$appear? (e) What is the greatest angle at which the light at wavelength$\mathbf{460}\mathbf{.}\mathbf{0}\mathbf{}\mathit{n}\mathit{m}\mathbf{}$appears?

The wavelength consists by light beam from $460nm$to $640nm$.

The diffraction grading, $\frac{W}{N}=\frac{1}{160lines/mm}$

The expression for the maxima of the diffraction is given as follows.

$\mathrm{dsin\theta }=\mathrm{m\lambda }\dots \dots \left(\mathrm{i}\right)$
The expression to calculate the angle is given as follows.

$\mathrm{\theta }={\sin }^{-1}\left(\frac{\mathrm{m\lambda }}{\mathrm{d}}\right)\dots \dots .\left(\mathrm{ii}\right)$

(a)

The width can be calculated as,

$d=\frac{W}{N}$

$d=\frac{1}{160lines/mm}$

$d=\frac{1}{160\times {10}^{-3}lines/m}$

$d=6.25\times {10}^{-6}m$

Calculate the angle for the wavelength$\lambda =460\text{nm}$and$m=2$.

Substitute for $460nm$for $\lambda$, $2$for $m$ and $6.25\times {10}^{-6}\text{m}$and for $d$into equation (ii).

$$\begin{gathered} \theta ={\sin }^{-1}\left(\frac{2\times 460\times 0^{-9}}{6.25\times {10}^{-6}}\right) \\ \theta ={\sin }^{-1}\left(0.1472\right) \\ \theta =8.46° \end{gathered}$$

Calculate the angle for the wavelength $\lambda =640\text{nm}$and $m=2$.

Substitute $640nm$for $\lambda$,$2$for $m$and $6.25\times {10}^{-6}\text{m}$for $d$into equation (ii).

$$\begin{gathered} \theta ={\sin }^{-1}\left(\frac{2\times 640\times 0^{-9}}{6.25\times {10}^{-6}}\right) \\ \theta ={\sin }^{-1}\left(0.2048\right) \\ \theta =11.82° \end{gathered}$$

Calculate the angle for the wavelength$\lambda =460\text{nm}$ and $m=3$.

Substitute $460nm$for $\lambda$,$3$for $m$and$6.25\times {10}^{-6}\text{m}$for $d$into equation (ii).

$$\begin{gathered} \theta ={\sin }^{-1}\left(\frac{3\times 460\times 0^{-9}}{6.25\times {10}^{-6}}\right) \\ \theta ={\sin }^{-1}\left(0.2208\right) \\ \theta =12.75° \end{gathered}$$

Calculate the angle for the wavelength$\lambda =640nm$ and $m=3$.

Substitute$640nm$for $\lambda$, $3$for $m$and$6.25\times {10}^{-6}\text{m}$for $d$into equation (ii).

$$\begin{gathered} \theta ={\sin }^{-1}\left(\frac{3\times 640\times 0^{-9}}{6.25\times {10}^{-6}}\right) \\ \theta ={\sin }^{-1}\left(0.3072\right) \\ \theta =17.89° \end{gathered}$$

Calculate the angle for the wavelength$\lambda =460nm$and $m=4$.

Substitute $460nm$for $\lambda$,$4$for $m$and $6.25\times {10}^{-6}\text{m}$for $d$into equation (ii).

$$\begin{gathered} \theta ={\sin }^{-1}\left(\frac{4\times 460\times 0^{-9}}{6.25\times {10}^{-6}}\right) \\ \theta ={\sin }^{-1}\left(0.2944\right) \\ \theta =17.12° \end{gathered}$$

Calculate the angle for the wavelength$\lambda =640nm$ and $m=4$.

Substitute $640nm$for $\lambda$, $4$for $m$and $6.25\times {10}^{-6}\text{m}$for $d$into equation (ii).

$$\begin{gathered} \theta ={\sin }^{-1}\left(\frac{4\times 640\times 0^{-9}}{6.25\times {10}^{-6}}\right) \\ \theta ={\sin }^{-1}\left(0.4096\right) \\ \theta =24.17° \end{gathered}$$

Since from the above calculation, it can be seen that the value of $\theta$the for $\lambda =460\text{nm}$,$m=4$ and $\lambda =640nm$,$m=3$ is nearly equals and overlaps. Therefore, the lowest order is $m=3$.

Hence the lowest order that is overlaps by the another is $3$.

(b)

To present the full spectral, the angle with the axis,$\theta =90°$

Substitute$90°$for $\theta$, $6.25\times {10}^{-6}\text{m}$for $d$and $640nm$for$\lambda$into equation (i).

$$\begin{gathered} 6.25\times {10}^{-6}\times \sin 90°=m\times 640\times {10}^{-9} \\ m=\frac{6.25\times {10}^{-6}\times \sin 90}{640\times {10}^{-9}} \\ m=\frac{6.25\times {10}^3}{640} \\ m=9.76 \end{gathered}$$

Hence the highest order at which full spectral can be seen is$9$ .

(c)

Calculate the value of angle,

Substitute $460nm$ for $\lambda$,$9$for $m$and $6.25\times {10}^{-6}\text{m}$ for $d$into equation (ii).

$$\begin{gathered} \theta ={\sin }^{-1}\left(\frac{9\times 460\times {10}^{-9}}{6.25\times {10}^{-6}}\right) \\ \theta ={\sin }^{-1}\left(0.6624\right) \\ \theta =41.48° \end{gathered}$$

Hence the angle at the highest order for wavelength$460nm$ is$41.48°$ .

(d)

Calculate the value of angle,

Substitute $640nm$for $\lambda$,$9$for$m$and $6.25\times {10}^{-6}\text{m}$ for $d$into equation (ii).

$$\begin{gathered} \theta ={\sin }^{-1}\left(\frac{9\times 640\times {10}^{-9}}{6.25\times {10}^{-6}}\right) \\ \theta ={\sin }^{-1}\left(0.9216\right) \\ \theta =67.16° \end{gathered}$$

Hence the angle at the highest order for wavelength$640nm$ is $67.16°$.

(e)

Calculate the highest order for which full spectral can be seen.

Substitute $90°$for $\theta$, $6.25\times {10}^{-6}\text{m}$ for $d$and $460nm$for $\lambda$ into equation (i).

$$\begin{gathered} 6.25\times {10}^{-6}\times \sin 90°=m\times 460\times {10}^{-9} \\ m=\frac{6.25\times {10}^{-6}\times \sin 90}{460\times {10}^{-9}} \\ m=\frac{6.25\times {10}^3}{460} \\ m=13.58 \end{gathered}$$

Calculate the value of angle,

Substitute $460nm$for $\lambda$, $13$for $m$and $6.25\times {10}^{-6}\text{m}$
for $d$into equation (ii).

$$\begin{gathered} \theta ={\sin }^{-1}\left(\frac{13\times 460\times {10}^{-9}}{6.25\times {10}^{-6}}\right) \\ \theta ={\sin }^{-1}\left(0.9568\right) \\ \theta =73.09° \end{gathered}$$

Hence the greatest angle which the light at wavelength $460nm$appears is$73.09°$ .