Chapter 24, Problem 26P

To determine

The maximum slit width.

Answer to Problem 26P

  $d_{\max }=\lambda$

Explanation of Solution

Given info:

Wavelength, $\lambda$

Formula used:

  $\sin \theta =\frac{m\lambda }{d}$

Where,

  $m$ is an integer

  $\lambda$ is wavelength

  $d$ is width

Calculation:

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

Rearrange the given formula: $\sin \theta =\frac{m\lambda }{d}$ . So, $d\sin {\theta }_{1\min }=m\lambda$

There will be no diffraction minima if the angle for the first minimum is greater than $90°$ .

Therefore, the limiting condition is

  $d_{\max }\sin \left(90°\right)=\left(1\right)\lambda$

  $d_{\max }=\lambda \left[\because \sin 90°=1\right]$

Conclusion:

The maximum width is $d_{\max }=\lambda$ .