91Ó°ÊÓ

The wavelengths of two emission are$\mathrm{λ}$ and $\mathrm{λ}+∆\mathrm{λ}$.

The angular separation is$∆\mathrm{θ}$ .

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.

$\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{θ}\mathbf{=}\mathit{m}\mathit{λ}$

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

Consider the diffraction grating for the emission which has wavelength $\mathrm{λ}$and corresponding angle is$\mathrm{θ}$.

…… (i)

$d\sin \left(\mathrm{θ}\right)=m\mathrm{λ}$

Consider the diffraction grating for the emission which has wavelength$\mathrm{λ}+∆\mathrm{λ}$and corresponding angle is$\mathrm{θ}+∆\mathrm{θ}$

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

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

From the definition of the derivative of$\sin \left(\mathrm{θ}\right)$,

$$\begin{gathered} \underset{∆\mathrm{θ}→∞}{lim}\sin \left(∆\mathrm{θ}\right)=\frac{\sin \left(\mathrm{θ}+∆\mathrm{θ}\right)-\sin \left(\mathrm{θ}\right)}{∆\mathrm{θ}} \\ \cos \mathrm{θ}=\frac{\sin \left(\mathrm{θ}+∆\mathrm{θ}\right)-\sin \left(\mathrm{θ}\right)}{∆\mathrm{θ}} \\ ∆\mathrm{θ}\cos \mathrm{θ}=\sin \left(\mathrm{θ}+∆\mathrm{θ}\right)-\sin \left(\mathrm{θ}\right) \end{gathered}$$

Substitute $∆\mathrm{θ}\cos \mathrm{θ}$for$\sin (\mathrm{θ}+∆\mathrm{θ})-\sin \left(\mathrm{θ}\right)$into equation (iii).

$$\begin{gathered} d∆\mathrm{θ}\cos \mathrm{θ}=m∆\mathrm{λ} \\ \mathrm{θ}∆=\frac{m∆\mathrm{λ}}{d\cos \mathrm{θ}} \end{gathered}$$

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

$∆\mathrm{θ}=\frac{m∆\mathrm{λ}}{d\sqrt{1-{\sin }^2\left(\mathrm{θ}\right)}}$

Substitute $m\mathrm{λ}/d$for $\sin \mathrm{θ}$into above equation.

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

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