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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 # How many lines per centimeter are there on a diffraction grating that gives a first-order maximum for ${\mathbf{470}}{\mathbf{‐}}{\mathbf{nm}}$ 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.

See the step by step solution

## Step 1: Given data

$\mathrm{\lambda }=470\mathrm{nm}\left(\frac{{10}^{‐9}\mathrm{m}}{1\mathrm{nm}}\right)=4.{70}^{,}{10}^{‐7}\mathrm{m}$

The wavelength of the blue light is

The order of maximum is 1.

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

## Step 2: Evaluating the lines per centimeter for first-order maximum

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,

$\mathrm{n}=\left(\frac{1.0\mathrm{cm}}{1.12×{10}^{‐6}\mathrm{m}}\right)\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)\phantom{\rule{0ex}{0ex}}\approx 8929$

Therefore, the lines for the first-order maximum are ${}^{\mathbf{8929}}$ per centimeter. ### Want to see more solutions like these? 