How many orders of the entire visible spectrum $\left(400-700\right)\mathbf{}\mathit{n}\mathit{m}$ can be produced by a grating of$\mathbf{500}\mathbf{}\raisebox{1ex}{$\mathbf{l}\mathbf{i}\mathbf{n}\mathbf{e}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{m}\mathbf{m}$}\right.$ ?

The wavelength of the spectrum $\lambda =700n\mathrm{m}$.

Diffraction grating,$\frac{W}{N}=\frac{1}{5\mathrm{li}ne/\mathrm{mm}}$

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

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

Here, $\mathit{d}$is the slit separation, $\mathit{m}$is the order of the diffraction,$\mathit{\lambda }$ is the wavelength and $\mathit{\theta }$is the angle.

The value of angle for which entire spectrum would exist is ${90}^{°}$and wavelength is$700nm$ .

Calculate the slit separation,

role="math" localid="1663138744202" $\begin{array}{l}d=\frac{w}{N}\\ d=\frac{1}{500x{10}^{3}line/m}\\ d=2\times {10}^{-6}m\end{array}$

Substitute the values and solve as:

$\begin{array}{l}d=\frac{1}{500x{10}^{3}line/m}\\ d=2\times {10}^{-6}m\end{array}$

Calculate the order of the entire visible spectrum.

Substitute $700nm$for $\lambda$, ${90}^{°}$for$\theta$ and$2\times {10}^{-6}$ for $d$into equation (i).

$$\begin{gathered} 2\times {10}^{-6}\times \sin {90}^{°}=m\times 700\times {10}^{-9} \\ m=\frac{2\times {10}^{-6}\times \sin {90}^{°}}{700\times {10}^{-9}} \\ m=\frac{2\times {10}^{-6}}{7\times {10}^{-7}} \\ m=2.876 \end{gathered}$$

Hence the maximum order for which full spectrum is exit is $2$.