91Ó°ÊÓ

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.

$\mathrm{»å²õ¾\pm ²Ôθ}=\mathrm{³¾Î»}……\left(\mathrm{i}\right)$
The expression to calculate the angle is given as follows.

$\mathrm{θ}={\sin }^{-1}\left(\frac{\mathrm{³¾Î»}}{\mathrm{d}}\right)…….\left(\mathrm{ii}\right)$

(a)

The width can be calculated as,

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

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

$d=\frac{1}{160×{10}^{-3}lines/m}$

$d=6.25×{10}^{-6}m$

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

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

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

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

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

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

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

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

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

Calculate the angle for the wavelength$\mathrm{λ}=640nm$ and $m=3$.

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

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

Calculate the angle for the wavelength$\mathrm{λ}=460nm$and $m=4$.

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

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

Calculate the angle for the wavelength$\mathrm{λ}=640nm$ and $m=4$.

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

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

Since from the above calculation, it can be seen that the value of $\mathrm{θ}$the for $\mathrm{λ}=460\text{nm}$,$m=4$ and $\mathrm{λ}=640nm$,$m=3$ 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,$\mathrm{θ}=90°$

Substitute$90°$for $\mathrm{θ}$, $6.25×{10}^{-6}\text{m}$for $d$and $640nm$for$\mathrm{λ}$into equation (i).

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

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

(c)

Calculate the value of angle,

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

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

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

(d)

Calculate the value of angle,

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

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

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 $\mathrm{θ}$, $6.25×{10}^{-6}\text{m}$ for $d$and $460nm$for $\mathrm{λ}$ into equation (i).

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

Calculate the value of angle,

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

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

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