Chapter 24, Problem 21P

To determine

The slit width.

Answer to Problem 21P

  $3.6\mu m$

Explanation of Solution

Given info:

Wavelength, $\lambda =653nm=653\times {10}^{-9}m$

Angle, $\theta =32°$

Formula used:

  $\sin \theta =\frac{m\lambda }{d}$

Where,

  $m$ is an integer

  $\lambda$ is wavelength

  $d$ is a width

Calculation:

The path difference is an integral multiple of the wavelength in constructive interference.

Rearrange the given formula: $\sin \theta =\frac{m\lambda }{d}$ .

The angle from the central maximum to the first bright fringe, $\theta =16°$

For constructive interference from the single slit, the path difference is $d\sin \theta =\left(m+\frac{1}{2}\right)\lambda ,m=1,2,3,\mathrm{...}$

For the first fringe away from the central maximum:

  $d\sin \left(16°\right)=\left(\frac{3}{2}\right)\left(653\times {10}^{-9}m\right)$

This gives $d=3.6\times {10}^{-6}m=3.6\mu m$

Conclusion: The required width is $3.6\mu m$ .