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Found in: Page 997

College Physics (Urone)

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

What is the highest-order maximum for 400-nm light falling on double slits separated by 25.0 µm?

The Highest-order maximum for 400-nm light falling on double slits separated by 25.0 µm is${}^{62}$.

See the step by step solution

Step 1: Given data

The wavelength of the violet light is$\mathrm{\lambda }=400\mathrm{nm}\left(\frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}\right)=4.0×{10}^{-7}\mathrm{m}$

The separation between the two slits is$\mathrm{d}=25\mathrm{\mu m}\left(\frac{{10}^{-6}\mathrm{m}}{1\mathrm{nm}}\right)=2.5×{10}^{-5}\mathrm{m}$

Step 2: Finding the highest-order maximum

A formula for the angle of the ${\mathrm{m}}^{\mathrm{th}}$-order maximum can be expressed as,

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

The highest order occurs when ${}^{\mathrm{sin}\mathrm{\theta }=1}$.

Therefore, the substitute for these values in equation (1) will give,

$\mathrm{d}=\mathrm{m\lambda }\phantom{\rule{0ex}{0ex}}\mathrm{m}=\frac{\mathrm{d}}{\mathrm{\lambda }}\phantom{\rule{0ex}{0ex}}\mathrm{m}=\frac{2.5\text{'}{10}^{-5}\mathrm{m}}{4.0\text{'}{10}^{-7}\mathrm{m}}\phantom{\rule{0ex}{0ex}}\mathrm{m}=62.5$

Therefore, we will get a maximum order of ${}^{\mathrm{m}:62}$