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Q102P

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Found in: Page 1080

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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# In a phasor diagram for any point on the viewing screen for the two slit experiment in Fig 35-10, the resultant wave phasor rotates${\mathbf{60}}{\mathbf{.}}{\mathbf{0}}{\mathbf{°}}$ in ${\mathbf{2}}{\mathbf{.}}{\mathbf{50}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{16}}{\mathbf{ }}{\mathbf{\text{s}}}$ . What is the wavelength?

Thus, the wavelength of light is $450nm$.

See the step by step solution

## Step 1: According to the question.

It is given that

$\Delta \varphi =60°\phantom{\rule{0ex}{0ex}}=\frac{\pi }{3}\text{rad}\phantom{\rule{0ex}{0ex}}$

The figure according to the question is:

The phasors rotate with constant angular velocity

${\mathbf{\omega }}{\mathbf{=}}\frac{\mathbf{\Delta \varphi }}{\mathbf{\Delta t}}\phantom{\rule{0ex}{0ex}}=\frac{\pi }{3×2.5×{10}^{-16} \text{s}}\phantom{\rule{0ex}{0ex}}=4.19×{10}^{15} \text{rad/s}\phantom{\rule{0ex}{0ex}}$

## Step 2: The wavelength of light in medium.

Since, light waves travelling in a medium (air) where the wave speed is approximately ${\mathbit{c}}$, then ${\mathbit{k}}{\mathbit{c}}{\mathbf{=}}{\mathbit{\omega }}{\mathbf{,}}{\mathbf{\text{where}}}{\mathbf{}}{\mathbit{k}}{\mathbf{=}}\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{\lambda }}$, which leads to

$\lambda =\frac{2\pi c}{\omega }\phantom{\rule{0ex}{0ex}}=\frac{2\mathrm{\pi }×3×{10}^{8}\mathrm{m}/\mathrm{s}}{4.19×{10}^{15}rad/s}\phantom{\rule{0ex}{0ex}}=4.498×{10}^{-7}m\phantom{\rule{0ex}{0ex}}=450nm$

Hence, the wavelength of light is, $450nm$.

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