Chapter 24, Problem 73GP

To determine

The linear distance between the first-orders fringes.

Answer to Problem 73GP

The separation of the lines is $0.7\text{m}$ .

Explanation of Solution

Given data:

Number of lines, $\frac{1}{\text{d}}=8600\text{lines/cm}$

Wavelength, ${\lambda }_1=4.6\times {10}^{-7}\text{m}$

Wavelength, ${\lambda }_2=6.8\times {10}^{-7}\text{m}$

Screen’s length, $\text{L}=2.5\text{m}$

Formula used:

  $d\sin \theta =m\lambda$ and $y=\text{L}\tan \theta$

Where,

  $L$ is the screen’s length.

  $\lambda$ is wavelength.

  $y$ is the distance from the white line on the screen.

  $d$ is the width of slit.

  $\theta$ is an angle.

Calculation:

The angle for the first-order from, $d\sin \theta =m\lambda$ ,

  $\left\{\frac{1}{\text{d}}=8600\text{lines/cm}\right\}\left({10}^{-2}\text{m/cm}\right)\sin {\theta }_1=4.6\times {10}^{-7}\text{m}$

  $\therefore \sin {\theta }_1=0.396$

  $\therefore {\theta }_1=23.3°$

Similarly,

  $\left\{\frac{1}{\text{d}}=8600\text{lines/cm}\right\}\left({10}^{-2}\text{m/cm}\right)\sin {\theta }_2=6.8\times {10}^{-7}\text{m}$

  $\therefore \sin {\theta }_2=0.585$

  $\therefore {\theta }_2=35.8°$

Now, the distance from the central white line on the screen: $y=\text{L}\tan \theta$

  $y_1=\text{L}\tan {\theta }_1=\left(2.5\text{m}\right)\tan 23.3°=1.1\text{m}$

  $y_2=\text{L}\tan {\theta }_2=\left(2.5\text{m}\right)\tan 35.8°=1.8\text{m}$

Therefore, the separation of the lines: $y_2-y_1$

  $y_2-y_1=\left(1.8-1.1\right)\text{m}=0.7\text{m}$

Conclusion: The separation of the lines is $0.7\text{m}$ .