Chapter 24, Problem 38P

To determine

The wavelengthsof $+26°38\text{'},+41°08\text{'}and-26°48\text{'},-41°19\text{'}$ .

Answer to Problem 38P

First wavelength is $473nm$ and second wavelength is $694nm$ .

Explanation of Solution

Given info:

Number of grating lines per cm, $N=9500lines/cm$

The first order spectral lines at $+26°38\text{'},+41°08\text{'}and-26°48\text{'},-41°19\text{'}$

Formula used:

  $\sin \theta =\frac{m\lambda }{d}$

Where,

  $m$ is an integer

  $\lambda$ is wavelength

  $d$ is distance between slits

Calculation:

Because the angles on each side of the central line are not the same, the incident light is not normal to the grating.

Now, calculate the average angles:

  ${\theta }_1=\frac{\left(26°38\text{'}+26°48\text{'}\right)}{2}=26°43\text{'}=26.72°$

  ${\theta }_2=\frac{\left(41°08\text{'}+41°19\text{'}\right)}{2}=41°14\text{'}=41.23°$

The path difference is an integer multiple of the wavelength in constructive interference.

Rearrange the given formula: $\sin \theta =\frac{m\lambda }{d}$ . So, $d\sin \theta =m\lambda$

Now, calculate the wavelengths by using $d\sin \theta =m\lambda$

For angle $\theta =26.72°$

  $\left[\frac{1}{\left(9500lines/cm\right)}\right]\left({10}^{-2}m/cm\right)\sin \left(26.72°\right)=\left(1\right){\lambda }_1$${\lambda }_1=4.73\times {10}^{-7}m=473nm$

For angle $\theta =41.23°$

  $\left[\frac{1}{\left(9500lines/cm\right)}\right]\left({10}^{-2}m/cm\right)\sin \left(41.23°\right)=\left(1\right){\lambda }_2$${\lambda }_1=6.94\times {10}^{-7}m=694nm$

Conclusion:

The required wavelengths: First wavelength is $473nm$ and second wavelength is $694nm$ .