Physics: Principles with Applications (7th Edition)

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

Chapter 25 - Optical Instruments - Problems - Page 741: 57

Answer

$1.0\times10^4 m$.

Work Step by Step

The angular half-width of the flashlight beam's central maximum, shining through an aperture D, is given in the middle of page 729, the same as equation 25–7. $$\theta=\frac{1.22\lambda}{D}$$ $$=\frac{1.22(550\times 10^{-9}m) }{(0.050m)}=1.3\times10^{-5}\;rad$$ By the definition of the radian, the radius of the circular spot when the beam strikes the moon is $\theta L$, where L is the distance from the earth to the moon. The diameter d of the beam when it reaches the moon is two times the radius of the circular spot. $$d=2(\theta L)=2(1.3\times10^{-5}\;rad)(3.84\times10^8 m)= 1.0\times10^4 m $$
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