Two emission lines have wavelengths$\mathit{\lambda }$ and $\mathit{\lambda }\mathbf{+}\mathbf{∆}\mathit{\lambda }$, respectively, where$\mathbf{∆}\mathit{\lambda }\mathbf{<}\mathit{\lambda }$ . Show that their angular separation$\mathbf{∆}\mathit{\theta }$ in a grating spectrometer is given approximately by

$\mathbf{∆}\mathit{\theta }\mathbf{=}\frac{\mathbf{∆}\mathbf{\lambda }}{\sqrt{\mathbf{(}\mathbf{d}\mathbf{/}\mathbf{m}{\mathbf{)}}^{\mathbf{2}}\mathbf{-}{\mathbf{\lambda }}^{\mathbf{2}}}}$

where $\mathit{d}$is the slit separation and$\mathit{m}$ is the order at which the lines are observed? Note that the angular separation is greater in the higher orders than the lower orders.

The wavelengths of two emission are$\lambda$ and $\lambda +∆\lambda$.

The angular separation is$∆\theta$ .

The slit separation is $d$and$m$ is the order of the diffraction.

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

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

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.

Consider the diffraction grating for the emission which has wavelength $\lambda$and corresponding angle is$\theta$.

…… (i)

$d\sin \left(\theta \right)=m\lambda$

Consider the diffraction grating for the emission which has wavelength$\lambda +∆\lambda$and corresponding angle is$\theta +∆\theta$

Subtract the equation (i) from the equation (ii).

$$\begin{gathered} \begin{array}{l}d\sin (\theta +∆\theta \}-d\sin \theta =m\left(\lambda +∆\lambda \right)-m\lambda \\ d\left[\sin \left(\theta +∆\theta \right)-\sin \theta \right]=m\lambda +m∆\lambda -m\lambda \end{array} \\ d\left[\sin \left(\theta +∆\theta \right)-\sin \theta \right]=m∆\lambda \end{gathered}$$ …… (iii)

From the definition of the derivative of$\sin \left(\theta \right)$,

$$\begin{gathered} \underset{∆\theta \to \infty }{lim}\sin \left(∆\theta \right)=\frac{\sin \left(\theta +∆\theta \right)-\sin \left(\theta \right)}{∆\theta } \\ \cos \theta =\frac{\sin \left(\theta +∆\theta \right)-\sin \left(\theta \right)}{∆\theta } \\ ∆\theta \cos \theta =\sin \left(\theta +∆\theta \right)-\sin \left(\theta \right) \end{gathered}$$

Substitute $∆\theta \cos \theta$for$\sin (\theta +∆\theta )-\sin \left(\theta \right)$into equation (iii).

$$\begin{gathered} d∆\theta \cos \theta =m∆\lambda \\ \theta ∆=\frac{m∆\lambda }{d\cos \theta } \end{gathered}$$

Substitute$\sqrt{1-{\sin }^2\left(\theta \right)}$for$\cos \theta$into above equation.

$∆\theta =\frac{m∆\lambda }{d\sqrt{1-{\sin }^2\left(\theta \right)}}$

Substitute $m\lambda /d$for $\sin \theta$into above equation.

role="math" localid="1663136987107" $$\begin{gathered} ∆\theta =\frac{m∆\lambda }{d\sqrt{1-{\left({\displaystyle \raisebox{1ex}{$m\lambda $}\!\left/ \!\raisebox{-1ex}{$d$}\right.}\right)}^2}} \\ ∆\theta =\frac{m∆\lambda }{\sqrt{d^2-{\left({\displaystyle m\lambda }\right)}^2}} \\ ∆\theta =\frac{m∆\lambda }{m\sqrt{{\left({\displaystyle \raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$m$}\right.}\right)}^2-{\left({\displaystyle \lambda }\right)}^2}} \\ ∆\theta =\frac{∆\lambda }{\sqrt{{\left({\displaystyle \raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$m$}\right.}\right)}^2-{\left({\displaystyle \lambda }\right)}^2}} \end{gathered}$$

Hence the required equation for the for angular separation in the grating spectrometers is$∆\theta =\frac{∆\lambda }{\sqrt{{\left({\displaystyle \raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$m$}\right.}\right)}^2-{\left({\displaystyle \lambda }\right)}^2}}$ .