Chapter 24, Problem 71GP

To determine

To Show: The second and third-order spectra overlap.

To Calculate: The exact wavelengths that overlap.

Answer to Problem 71GP

  $600\text{nm}$ to $750\text{nm}$ of the second-order overlaps with $400\text{nm}$ to $500\text{nm}$ of the third order.

Explanation of Solution

Given data:

It is given that there is second-order spectrum overlap with third order spectrum.

Formula used:

  $\sin \theta =\frac{m\lambda }{\text{d}}$

Where,

  $d$ is the distance between lines.

  $\lambda$ is wavelength.

  $\theta$ is angle of the normal.

Calculation:

The angle increases the wavelength increases.

Comparing the maximum angle for the second-order with the minimum angle for third-order,

  $\sin \theta =\frac{m\lambda }{\text{d}}$

  $\sin {\theta }_{2\max }=\frac{2\left(750\text{nm}\right)}{\text{d}}$ and $\sin {\theta }_{3\min }=\frac{3\left(400\text{nm}\right)}{\text{d}}$

Dividing the both equations,

  $\frac{\sin {\theta }_{3\min }}{\sin {\theta }_{2\max }}=\frac{1200\text{nm}}{1500\text{nm}}=0.8$

As the value of sine increases with angle, i.e. ${\theta }_{3\min }<{\theta }_{2\max }$ , so the orders overlap.

Let the second-order wavelength coincides with ${\theta }_{3\min }$ to determine the overlap:

  $\left(2\right){\lambda }_2=\left(3\right)\left(400\text{nm}\right)$

  $\therefore {\lambda }_2=600\text{nm}$

Let the third-order wavelength coincides with ${\theta }_{2\max }$ to determine the overlap:

  $\left(2\right)\left(750\text{nm}\right)=\left(3\right){\lambda }_3$

  $\therefore {\lambda }_3=500\text{nm}$

Conclusion: $600\text{nm}$ to $750\text{nm}$ of the second-order overlaps with $400\text{nm}$ to $500\text{nm}$ of the third order.