Chapter 24, Problem 23P

To determine

The width of the slit.

Answer to Problem 23P

  $9.53\times {10}^{-7}m$

Explanation of Solution

Given info:

Wavelength, $\lambda =440nm=440\times {10}^{-9}m$

Angle, $2\theta =55.0°\Rightarrow \theta =\frac{55°}{2}=27.5°$

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.

Find the angular half-width $\theta$ of the central maximum from $\sin \theta =\frac{\lambda }{d}$

  $\sin \left(\frac{55.0°}{2}\right)=\frac{440\times {10}^{-9}m}{d}$

  $d=9.53\times {10}^{-7}m$

Conclusion: The required width is $9.53\times {10}^{-7}m$ .