Chapter 20, Problem 20.AE

(a)

Interpretation Introduction

Interpretation:

The wavelength of two spectral lines has to be distinguished.

Concept introduction:

The wavelength of two spectral lines calculated via below the formula

$\begin{array}{l}\frac{\text{λ}}{\text{Δλ}}\\ \text{where}\\ \text{λ}\text{=}\text{wavelength}\text{of spectral line}\\ \end{array}$

Answer to Problem 20.AE

The wavelength of two spectral lines will be resolved.

Explanation of Solution

Given

$\begin{array}{l}\text{λ=}\text{10}\text{.00}\text{μm}\\ \text{and}\text{10}\text{.01}\text{μm}\end{array}$

$\begin{array}{l}\text{the}\text{value}\text{of}\text{λ=}\text{10}\text{.00}\text{μm}\text{and}\text{Δλ0}\text{.01}\text{μm}\\ \frac{\text{λ}}{\text{Δλ}}\text{=}\frac{\text{10}\text{.00}}{\text{0}\text{.01}}\\ \text{=}{\text{10}}^{\text{3}}\end{array}$

The given resolution is ${\text{10}}^4$ so the observed and given both resolution almost equal so the lines will be resolved.

(b)

Interpretation Introduction

Interpretation:

The resolution close to wavenumber has to be has to be calculated.

Concept introduction:

The resolution close to wavenumber calculated via below the formula

$\begin{array}{l}\text{λ=}\frac{\text{1}}{\widetilde{\text{υ}}}\\ \text{where}\\ \text{λ}\text{=}\text{wavelength}\text{of spectral line}\\ \widetilde{\text{υ}}\text{=}\text{wave}\text{number}\end{array}$

Answer to Problem 20.AE

The resolution close to wavenumber both deference $\text{is}\text{0}\text{.1}{\text{cm}}^{\text{-1}}$

Explanation of Solution

The given resolution is ${\text{10}}^4$

$\text{λ=}\frac{\text{1}}{\widetilde{\text{υ}}}$

$\begin{array}{l}\text{λ=}\frac{\text{1}}{\text{(1000}{\text{cm}}^{\text{-1}}{\text{)(10}}^{\text{-4}}\text{cm/μm)}}\\ \text{=}\text{10μm}\\ \text{Δλ}\text{=}\frac{\text{λ}}{{\text{10}}^{\text{4}}}\\ \text{=}{\text{10}}^{\text{-3}}\text{μm}\end{array}$

$\text{10}\text{.001μm}\text{cound}\text{be }\text{resolved}\text{from}\text{100}\text{μm}$

$\begin{array}{l}\text{10}\text{.001μm}\text{=}\text{1000}{\text{cm}}^{\text{-1}}\\ \text{10}\text{.001μm}\text{=}\text{999}\text{.9}{\text{cm}}^{\text{-1}}\end{array}\}\text{thedifference}\text{is}\text{0}\text{.1}{\text{cm}}^{\text{-1}}$

(c)

Interpretation Introduction

Interpretation:

The resolution of first and tenth-order has to be calculated.

Answer to Problem 20.AE

The resolution of first and tenth-order is $\text{12500}\text{for}\text{n}\text{=1}$ and $\text{125000}\text{for}\text{n}\text{=10}$

Explanation of Solution

Given

$\text{5}\text{.0}\text{cm}\text{×}\text{2500}\text{line/cm=12500}\text{lines}$

Resolution:

$\begin{array}{l}\text{=1}\text{.12500}\\ \text{12500}\text{for}\text{n}\text{=1}\\ \text{=10}\text{.12500}\\ \text{125000}\text{for}\text{n}\text{=10}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The angular dispersion has to be fined.

Concept introduction:

Dispersion: It measure the ability to separate wavelengths differing by $\text{Δλ}$ through the difference in angle $\text{Δ}\varphi$

Dispersion of grating:

$\begin{array}{l}\frac{\text{Δ}\varphi }{\text{Δλ}}\text{=}\frac{\text{n}}{\text{d}\text{cos}\varphi }\\ \text{n=}\text{is}\text{the}\text{diffraction}\text{order}\end{array}$

Answer to Problem 20.AE

The angular dispersion is $\text{0}{\text{.3}}^{\text{o}}$

Explanation of Solution

$\frac{\text{Δ}\varphi }{\text{Δλ}}\text{=}\frac{\text{n}}{\text{d}\text{cos}\varphi }$

$\frac{\text{2}}{\left(\frac{\text{1}\text{mm}}{\text{250}}\right)\text{cos}{\text{30}}^{{}^{\text{o}}}}$

$\begin{array}{l}\text{577}\frac{\text{radians}}{\text{mm}}\text{=}\text{0}\text{.577}\frac{\text{radians}}{\text{μm}}\\ \text{convert}\text{radians}\text{to}\text{degrees}\frac{\text{radians}}{\text{π}}\text{×}\text{180}\end{array}$

$\frac{\text{Δf}}{\text{Δλ}}\text{=}\text{33}\text{.1}\text{degrees/μm}$

Two wavelengths are

$\begin{array}{l}\text{1000}{\text{cm}}^{\text{-1}}\\ \text{=}\text{10}\text{μm}\text{and}{\text{1001cm}}^{\text{-1}}\\ \text{Δλ}\text{=}\text{0}\text{.01μm}\end{array}$

$\begin{array}{l}\text{Δ}\lambda \text{=0}\text{.577}\frac{\text{radians}}{\text{mm}}\text{×}\text{0}\text{.01μm}\\ \text{=}\text{6}\text{×}{\text{10}}^{\text{-3}}\text{radian}\\ \text{=}\text{0}{\text{.3}}^{\text{o}}\end{array}$