Chapter 24, Problem 22P

To determine

The width of the central diffraction peak.

Answer to Problem 22P

  $7.79cm$

Explanation of Solution

Given info:

Wavelength, $\lambda =589nm=589\times {10}^{-9}m$

Width of the slit, $d=0.0348mm=0.0348\times {10}^{-3}m$

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: $\sin \theta =\frac{m\lambda }{d}$ .

So, $\sin {\theta }_{1\min }=\frac{m\lambda }{d}=\frac{\left(1\right)\left(589\times {10}^{-9}m\right)}{\left(0.0348\times {10}^{-3}m\right)}=0.0169$

It gives, ${\theta }_{1\min }=0.970°$

The distance on the screen from

  $y_1=L\tan {\theta }_1=\left(2.30m\right)\tan 0.970°=3.89\times {10}^{-2}m=3.89cm$

The width of the peak is $\Delta y_1=2y_1=2\left(3.89cm\right)=7.79cm$

Conclusion: The width of the central peak is $7.79cm$ .