Suggested languages for you:

Americas

Europe

Q11PE

Expert-verified
Found in: Page 997

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

What is the wavelength of light falling on double slits separated by 2.00 µm if the third-order maximum is at an angle of 60.0º ?

The wavelength of the light is${}^{577\mathrm{nm}}$.

See the step by step solution

Step 1: Given data

The angle for the third minimum is${}^{\mathbf{60}\mathbf{.}{\mathbf{0}}^{\mathbf{0}}}$

The separation of the double slits is $\mathrm{d}=2.00\mathrm{\mu m}\left(\frac{{10}^{-6}\mathrm{m}}{1\mathrm{\mu m}}\right)=2.00×{10}^{-6}\mathrm{m}$

Step 2: Finding the wavelength of the light

A formula for the angle of the third-order maximum can be expressed as,

$\mathrm{dsin\theta }=3\mathrm{\lambda }...............................\left(1\right)$

Substituting the given data in equation (1), we get,

$2.00×{10}^{-6}\mathrm{m}×\mathrm{sin}\left(60.{0}^{0}\right)=3×\mathrm{\lambda }\phantom{\rule{0ex}{0ex}}\mathrm{\lambda }=\left(\frac{1.732×{10}^{-6}\mathrm{m}}{3}\right)\phantom{\rule{0ex}{0ex}}\mathrm{\lambda }=0.577×{10}^{-6}\mathrm{m}\left(\frac{1\mathrm{nm}}{{10}^{-9}\mathrm{m}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{\lambda }=577\mathrm{nm}$

Thus, the distance between two slits is ${}^{\mathbf{577}\mathbf{}\mathbf{nm}}$.