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Q23PE

Expert-verifiedFound in: Page 997

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**How many lines per centimeter are there on a diffraction grating that gives a first-order maximum for ${\mathbf{470}}{\mathbf{\u2010}}{\mathbf{nm}}$ ****blue light at an angle of **${\mathbf{25}}{\mathbf{.}}{{\mathbf{0}}}^{{\mathbf{\circ}}}$**?**

The lines per cm for the first-order maximum are $8929$ per centimeter.

$\mathrm{\lambda}=470\mathrm{nm}\left(\frac{{10}^{\u20109}\mathrm{m}}{1\mathrm{nm}}\right)=4.{70}^{,}{10}^{\u20107}\mathrm{m}$

The wavelength of the blue light is

The order of maximum is 1.

The angle at which the first-order maximum is observed ${}_{\mathbf{25}\mathbf{.}{\mathbf{0}}^{\mathbf{\circ}}}$

The equation of the constructive interference can be expressed as

$\mathrm{d}\mathrm{sin\theta}=\mathrm{m\lambda}$ …………………(1)

Where, $d,\lambda ,\theta $ and m is the separation between slits, the wavelength of the incident light, and the angle made with the maximum and order of maximum, respectively.

Therefore, if we substitute the given data in equation (1) for m=1, we will get,

We then evaluate the number of lines per centimeter, $\mathrm{n}$, using the relation:

$\mathrm{n}=\frac{1.0\mathrm{cm}}{\mathrm{d}}$.

Substitute the value in the above expression, and we get,

$\mathrm{n}=\left(\frac{1.0\mathrm{cm}}{1.12\times {10}^{\u20106}\mathrm{m}}\right)\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)\phantom{\rule{0ex}{0ex}}\approx 8929$

Therefore, the lines for the first-order maximum are ${}^{\mathbf{8929}}$ per centimeter.

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