Chapter 24, Problem 70GP

To determine

The number of lines per centimetre when there is no second-order spectrum visible.

Answer to Problem 70GP

  $12,500\text{lines/cm}$

Explanation of Solution

Given data:

It is given that there is no second-order spectrum visible.

Formula used:

  $d\sin \theta =m\lambda$

Where,

  $d$ is the distance between lines.

  $\lambda$ is wavelength.

  $\theta$ is angle of the normal.

Calculation:

The angle increases, the wavelength increases. But, it is given that there is no second-order spectrum visible, so use the smallest wavelength.

Take maximum angle of $90°$ .

Consider first order and find the slit separation using the formula, $d\sin \theta =m\lambda$

  $\therefore d\sin 90°=\left(2\right)\left(400\times {10}^{-9}\text{m}\right)$

  $d=8.00\times {10}^{-5}\text{cm}$

  $\frac{1}{d}=\frac{1}{8.00\times {10}^{-5}\text{cm}}=12,500\text{lines/cm}$

Conclusion: The number of lines are $12,500\text{lines/cm}$ .