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Q10PE

Expert-verifiedFound in: Page 997

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 ${}^{600\mathrm{nm}}$.

The angle for the third minimum is ${}^{30.0\xb0}$

The separation of the double slits is ${}^{d=3.00\mu m\left(\frac{{10}^{-\xb0}}{1\mu m}\right)=3.00\times {10}^{-6}m}$

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,

$3.00\times {10}^{-6}m\times \mathrm{sin}\left(30.0\xb0\right)=\frac{5}{2}\times \lambda \phantom{\rule{0ex}{0ex}}\lambda =\left(\frac{1.50\times {10}^{-6}m}{2.5}\right)\phantom{\rule{0ex}{0ex}}\lambda =0.600\times {10}^{-6}m\left(\frac{1nm}{{10}^{-9}m}\right)\phantom{\rule{0ex}{0ex}}\lambda =600nm$

Thus, the distance between two slits is ${}^{600\mathrm{nm}}$.

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