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Expert-verified Found in: Page 997 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # Calculate the wavelength of light that has its third minimum at an angle of 30.0º when falling on double slits separated by 3.00 µm.

The wavelength of the light is ${}^{600\mathrm{nm}}$.

See the step by step solution

## Step 1: Given data

The angle for the third minimum is ${}^{30.0°}$

The separation of the double slits is ${}^{d=3.00\mu m\left(\frac{{10}^{-°}}{1\mu m}\right)=3.00×{10}^{-6}m}$

## Step 2: Finding the wavelength of the light

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

$\mathrm{d}\mathrm{sin}\mathrm{\theta }=\frac{5}{2}\mathrm{\lambda }$………………………….(1)

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

$3.00×{10}^{-6}m×\mathrm{sin}\left(30.0°\right)=\frac{5}{2}×\lambda \phantom{\rule{0ex}{0ex}}\lambda =\left(\frac{1.50×{10}^{-6}m}{2.5}\right)\phantom{\rule{0ex}{0ex}}\lambda =0.600×{10}^{-6}m\left(\frac{1nm}{{10}^{-9}m}\right)\phantom{\rule{0ex}{0ex}}\lambda =600nm$

Thus, the distance between two slits is ${}^{600\mathrm{nm}}$. ### Want to see more solutions like these? 