Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 24 - The Wave Nature of Light - General Problems - Page 711: 85

Answer

$3.19\times10^{-5}m$.

Work Step by Step

We see that the angles making the central peak of the diffraction pattern are very small, because the width of the pattern is very small compared to the distance to the screen. For small angles, the sine and tangent functions are approximately equal. $$sin \theta \approx tan \theta $$ Calculate the angle to the first minimum, using geometry. The distance from the centerline to the first minimum is half the width of the full first-order pattern. $$tan \theta=\frac{0.5(8.20\times10^{-2}m)}{3.15m}=0.01302 = sin \theta$$ Apply equation 24–3b and solve for the slit width. $$D sin \theta = m \lambda$$ $$D=\frac{m \lambda}{sin \theta}=\frac{(1)(415nm)}{0.01302}=3.188\times10^4 nm=3.19\times10^{-5}m$$
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