Chapter 24, Problem 19P

To determine

The distance of first fringe from the strong central maximum.

Answer to Problem 19P

  $2.51m$

Explanation of Solution

Given info:

Wavelength, $\lambda =520nm=520\times {10}^{-9}m$

Distance, $d=3.20\times {10}^{-3}mm=3.20\times {10}^{-6}m$

Formula used:

It is known that $\sin \theta =\frac{m\lambda }{d}$

Where,

  $m$ is an integer

  $\lambda$ is wavelength

  $d$ is a distance

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, $d\sin \theta =m\lambda$

According to the question, 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,

  $\left(3.20\times {10}^{-6}m\right)\sin {\theta }_1=\left(\frac{3}{2}\right)\left(520\times {10}^{-9}m\right)$

This gives out: ${\theta }_1=14.1°$

Now, calculate the distance on the screen from

  $y_1=L\tan {\theta }_1=\left(10.0m\right)\tan 14.1°=2.51m$

Conclusion: The required value is $2.51m$ .