Chapter 24, Problem 8P

To determine

The angle that is relative to the “straight-through” direction would there be little or nil wave action.

Explanation of Solution

Given info:

Wavelength, $\lambda =2.5cm$

Distance, $d=5$

Order, $m=0,1,2,\mathrm{...}$

Formula used:

  $\sin \theta =\frac{m\lambda }{d}$

Where,

  $m$ is an integer

  $\lambda$ is wavelength

  $d$ is a distance

Calculation:

The equation can be set-up as $\sin \theta =\frac{\left(m+0.5\right)\left(2.5\right)}{5}=\left(m+0.5\right)0.5,m=0,1,2,\mathrm{...}$

Solve for $\theta$

  $m=0\to {\sin }^{-1}\left(\left(0+0.5\right)0.5\right)=14.4°$

  $m=1\to {\sin }^{-1}\left(\left(1+0.5\right)0.5\right)=48.5°$