Chapter 28, Problem 62GP

To determine

To Prove: The diffractive spread of the laser beam $\approx \frac{\lambda }{D}$ as described in section 28-11.

Explanation of Solution

Calculation:

The uncertainty in the momentum component perpendicular to the motion is,

  $\begin{array}{c}\Delta p_y\ge \frac{h}{\Delta y}\\ =\frac{\left(\frac{h}{2\pi }\right)}{D}\end{array}$

Consider the half angle of the beam is,

  $\sin \theta =\frac{v_y}{c}$

Consider if the angle is small $\theta$ then,

  $\begin{array}{c}\theta =\frac{v_y}{v_x}\\ =\frac{p_y}{p}\\ =\frac{\left(\frac{h}{2\pi }\right)}{\frac{h}{\lambda }}\\ =\frac{\lambda }{2\pi D}\end{array}$

Consider the angular speed is,

  $\begin{array}{l}2\theta =\frac{2\lambda }{\pi D}\\ \theta =\frac{\lambda }{\pi D}\end{array}$

Conclusion:

Therefore, the value of the angular speed is $\theta =\frac{\lambda }{\pi D}$ .