/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 76 A \(10-\) W laser emits a beam o... [FREE SOLUTION] | 91影视

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Chapter 22: Problem 76

A \(10-\) W laser emits a beam of light 4.0 mm in diameter. The laser is aimed at the Moon. By the time it reaches the Moon, the beam has spread out to a diameter of \(85 \mathrm{km} .\) Ignoring absorption by the atmosphere, what is the intensity of the light (a) just outside the laser and (b) where it hits the surface of the Moon?

Short Answer

Expert verified
Answer: The intensity of the light just outside the laser is \(7.95951 \times 10^5 \ \text{W/m}^2\), and the intensity of the light at the surface of the Moon is \(1.7622 \times 10^{-9} \ \text{W/m}^2\).

Step by step solution

01

Find the initial area of the laser beam just outside the laser.

To find the initial area of the laser beam, we'll use the formula for the area of a circle: \(A = \pi r^2\), where A is the area and r is the radius of the circle. The given diameter is 4.0 mm, so the radius is half that: \(r = \frac{4.0 \ \text{mm}}{2} = 2.0 \ \text{mm} = 2.0 \times 10^{-3} \ \text{m}\). Now we can find the initial area: \(A_1 = \pi (2.0 \times 10^{-3} \ \text{m})^2 = 1.2566 \times 10^{-5} \ \text{m}^2\).
02

Find the area of the laser beam at the surface of the Moon.

We're given that the diameter of the laser beam at the surface of the Moon is 85 km, so the radius is half that: \(r = \frac{85 \ \text{km}}{2} = 42.5 \ \text{km} = 42.5 \times 10^3 \ \text{m}\). Now we can find the area at the Moon's surface: \(A_2 = \pi (42.5 \times 10^3 \ \text{m})^2 = 5.6721 \times 10^{9} \ \text{m}^2\).
03

Calculate the intensity just outside the laser.

We can now use the power and initial area, \(A_1\), to find the intensity just outside the laser: \(I_1 = \frac{P}{A_1} = \frac{10 \ \text{W}}{1.2566 \times 10^{-5} \ \text{m}^2} = 7.95951 \times 10^5 \ \text{W/m}^2\).
04

Calculate the intensity at the surface of the Moon.

Similarly, we can use the power and area on the Moon's surface, \(A_2\), to find the intensity at the Moon's surface: \(I_2 = \frac{P}{A_2} = \frac{10 \ \text{W}}{5.6721 \times 10^{9} \ \text{m}^2} = 1.7622 \times 10^{-9} \ \text{W/m}^2\).
05

Write down the final answers.

The intensity of the light just outside the laser is \(7.95951 \times 10^5 \ \text{W/m}^2\), and the intensity of the light at the surface of the Moon is \(1.7622 \times 10^{-9} \ \text{W/m}^2\).

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