Chapter 19.2, Problem 29SSC

To determine

The separation of the first-order dark bands.

Answer to Problem 29SSC

  $0.46\text{cm}$

Explanation of Solution

Given:

The wavelength of the light, $\lambda =546\text{nm}$

The distance between the slits, $d=0.08\text{mm}$

The distance of the screen from the grating, $L=68.0\text{cm}$

Formula used:

Wavelength of light can be obtained by:

  $\lambda =d\sin \theta$

Where,

  $\theta$ is the angle at which the first order bright line occurs

d is the distance between the slits.

The angle at which the first-order bright line occurs can also be written as,

  $\theta ={\tan }^{-1}\left(\frac{x}{L}\right)$

Where,

x is the separation of the lines in the diffraction pattern

L is the distance between the grating and screen.

Calculation:

Now using the formula for the wavelength, the angle at which the first-order bright line occurs can be written as,

  $\begin{array}{c}\lambda =d\sin \theta \\ 546\times {10}^{-9}\text{m}=\left(0.08\times {10}^{-3}\right)\sin \theta \\ \theta =0.39°\end{array}$

Using the above and given values, the separation of the first-order dark bands will be,

  $\begin{array}{c}\tan \theta =\frac{x}{L}\\ \tan \left(0.39°\right)=\frac{x}{68.0\text{cm}}\\ x=0.46\text{cm}\end{array}$

Conclusion:

Hence, the separation of the first-order dark bands is $0.46\text{cm}$ .