## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 22 - Problems - Page 866: 71

#### Answer

(a) The intensity of the laser beam is $1.13\times 10^3~W/m^2$ (b) The intensity of the light incident on the retina is $6.37\times 10^6~W/m^2$ (c) The total energy incident on the retina is $1.6\times 10^{-4}~J$

#### Work Step by Step

(a) We can find the intensity of the laser beam: $I = \frac{P}{A}$ $I = \frac{P}{\pi~r^2}$ $I = \frac{2.0\times 10^{-3}~W}{(\pi)(0.75\times 10^{-3}~m)^2}$ $I = 1.13\times 10^3~W/m^2$ The intensity of the laser beam is $1.13\times 10^3~W/m^2$ (b) We can find the intensity of the light incident on the retina: $I = \frac{P}{A}$ $I = \frac{P}{\pi~r^2}$ $I = \frac{2.0\times 10^{-3}~W}{(\pi)(10.0\times 10^{-6}~m)^2}$ $I = 6.37\times 10^6~W/m^2$ The intensity of the light incident on the retina is $6.37\times 10^6~W/m^2$ (c) We can find the total energy incident on the retina: $E = P~t$ $E = (2.0\times 10^{-3}~W)(80\times 10^{-3}~s)$ $E = 1.6\times 10^{-4}~J$ The total energy incident on the retina is $1.6\times 10^{-4}~J$

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