Chapter 24, Problem 18P

To determine

The wavelength of the light used.

Answer to Problem 18P

  $782nm$

Explanation of Solution

Given info:

Angle, $\theta =35°$

Width of the slit, $d=2.6\times {10}^{-3}mm$

Formula used:

  $\sin \theta =\frac{m\lambda }{d}$

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

The angle from the central maximum to the first minimum is $17.5°$ .

Now, find the wavelength form substituting the values in $d\sin \theta =m\lambda$

  $\left(2.60\times {10}^{-6}m\right)\sin \left(17.5°\right)=\left(1\right)\lambda$

This gives out $\lambda =7.82\times {10}^{-7}m=782nm$

Conclusion: The required wavelength is $782nm$ .