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Q101P

Expert-verifiedFound in: Page 1080

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Find the slit separation of a double-slit arrangement that will produce interference fringes ${\mathbf{0}}{\mathbf{.}}{\mathbf{018}}{\mathbf{\u200a}}{\mathbf{\text{rad}}}$****apart on a distant screen when the light has wavelength ${\mathbf{\lambda}}{\mathbf{=}}{\mathbf{589}}{\mathbf{\u200a}}{\mathbf{\text{nm}}}$.**

Thus, the wavelength of visible light is$33\mu m$.

**In the case of a distant screen the angle** ${\mathbf{\theta}}$ ** is close to zero so ${\mathit{s}}{\mathit{i}}{\mathit{n}}{\mathit{\theta}}{\mathbf{\approx}}{\mathit{\theta}}$**

** **

**Thus, solve further gives:**

**${\Delta}{\theta}{\approx}{\Delta}{s}{i}{n}{\theta}\phantom{\rule{0ex}{0ex}}{=}{\Delta}\left(\frac{m\lambda}{d}\right)\phantom{\rule{0ex}{0ex}}{=}\frac{\lambda}{d}{\Delta}{m}\phantom{\rule{0ex}{0ex}}{\Delta}{\theta}{=}\frac{\lambda}{d}\phantom{\rule{0ex}{0ex}}$**

**Here, **** ${\mathit{\lambda}}$ is wavelength of light in air/vacuum.**

Use the above formula as follows:

$d=\frac{\lambda}{\Delta \theta}\phantom{\rule{0ex}{0ex}}=\frac{589\times {10}^{-9}m}{0.018rad}\phantom{\rule{0ex}{0ex}}=3.3\times {10}^{-5}m\phantom{\rule{0ex}{0ex}}=33\mu m\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Hence, the wavelength of visible light is$33\mu m$.

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