StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q102P

Expert-verifiedFound in: Page 1080

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**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${\mathbf{60}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\xb0}}$ in** ${\mathbf{2}}{\mathbf{.}}{\mathbf{50}}{\mathbf{\times}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{16}}{\mathbf{\u200a}}{\mathbf{\text{s}}}$ **. What is the wavelength?**

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

It is given that

$\Delta \varphi =60\xb0\phantom{\rule{0ex}{0ex}}=\frac{\pi}{3}\text{rad}\phantom{\rule{0ex}{0ex}}$

The figure according to the question is:

The phasors rotate with constant angular velocity

${\mathbf{\omega}}{\mathbf{=}}\frac{\mathbf{\Delta \varphi}}{\mathbf{\Delta t}}\phantom{\rule{0ex}{0ex}}=\frac{\pi}{3\times 2.5\times {10}^{-16}\u200a\text{s}}\phantom{\rule{0ex}{0ex}}=4.19\times {10}^{15}\u200a\text{rad/s}\phantom{\rule{0ex}{0ex}}$

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

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

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

94% of StudySmarter users get better grades.

Sign up for free