Question

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.

Short answer

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$}\right.}\right)}^2-{\lambda }^2}}$.

Step-by-step solution

Write the given data from the question.

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.

Determine the formulas to derive the expression for angular separation in the grating spectrometers.

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.

Derive the expression for angular separation in the grating spectrometers.

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$\theta +∆\theta$

$dsin(\theta +∆\theta )=m(\lambda +∆\lambda )$ …… (ii)

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$}\right.}\right)}^2-{\left({\displaystyle \lambda }\right)}^2}}$ .