Chapter 24, Problem 20P

To determine

The width of the central maximum (in cm) in the diffraction pattern on a screen that is $5.0m$ away.

Answer to Problem 20P

  $0.45cm$

Explanation of Solution

Given info:

Wavelength, $\lambda =450nm=450\times {10}^{-9}m$

Width of slit, $d=1.00mm=1.00\times {10}^{-3}m$

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 integer multiple of the wavelength in constructive interference.

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

According to the question, find the angle to the first minimum from $\sin {\theta }_{1\min }=\frac{m\lambda }{d}$

  $\sin {\theta }_{1\min }=\frac{m\lambda }{d}=\frac{\left(1\right)\left(450\times {10}^{-9}m\right)}{1.0\times {10}^{-3}m}=0.00045$

Now, calculate the distance from the screen using $y_1=L\tan {\theta }_1$

For small angle:

  $\sin \theta \approx \tan \theta$ this gives $y=L\sin \theta =\left(5.0m\right)\left(0.00045\right)=0.00225m$

Therefore, the width of the central maximum is $2y=0.0045m=0.45cm$

Conclusion: The width of the central maximumis $0.45cm$ .