Chapter 24, Problem 9P

To determine

To Describe: The interference pattern on the screen.

Explanation of Solution

Given info:

Angle, $\theta =180°$

Path length, $\frac{1}{2}\left(\lambda \right)$

Formula used:

  $\sin \theta =\frac{m\lambda }{d}$

Where,

  $m$ is an integer

  $\lambda$ is wavelength

  $d$ is a distance

Calculation:

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, constructive interference: $\frac{1}{2}\lambda +d\sin \left({\theta }_{\max }\right)=m\lambda$

  $m=0,\pm 1,\pm 2,\pm 3,\pm 4,\mathrm{...}$

  $d\sin \left(\theta \right)=\left(m-\frac{1}{2}\right)\lambda$

  $m=0,\pm 1,\pm 2,\pm 3,\pm 4,\mathrm{...}$

Now, destructive interference: $\frac{1}{2}\lambda +d\sin \left({\theta }_{\max }\right)=\left(m+\frac{1}{2}\right)\lambda$

  $m=0,\pm 1,\pm 2,\pm 3,\pm 4,\mathrm{...}$

  $d\sin \left(\theta \right)=m\lambda$

  $m=0,\pm 1,\pm 2,\pm 3,\pm 4,\mathrm{...}$

Therefore, pattern: reverse of usual double slit pattern.