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Expert-verified Found in: Page 1080 ### Fundamentals Of Physics

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 ${\mathbf{0}}{\mathbf{.}}{\mathbf{018}}{\mathbf{ }}{\mathbf{\text{rad}}}$apart on a distant screen when the light has wavelength ${\mathbf{\lambda }}{\mathbf{=}}{\mathbf{589}}{\mathbf{ }}{\mathbf{\text{nm}}}$.

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

See the step by step solution

## Step 1: According to the question.

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

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, ${\mathbit{\lambda }}$ is wavelength of light in air/vacuum.

## Step 2: The wavelength of visible light.

Use the above formula as follows:

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

Hence, the wavelength of visible light is$33\mu m$. ### Want to see more solutions like these? 