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Q13PE

Expert-verifiedFound in: Page 997

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${}^{62}$.

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

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

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 ${}^{\mathrm{m}:62}$

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