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Chapter 14: Problem 54

Light emerges from a 5.0 -mW laser in a beam 1.0 mm in diameter. The beam shines on a wall, producing a spot \(3.6 \mathrm{cm}\) in diameter. What is the beam's intensity (a) at the laser and (b) at the wall?

Short Answer

Expert verified
The beam's intensity is (a) approximately \(6.37 * 10^{3} \) W/m² at the laser and (b) approximately \(4.90 \) W/m² at the wall.

Step by step solution

01

Compute the intensity at the laser.

The power of the laser has been given as 5.0 milliwatts or \(5.0 * 10^{-3}\) watts. The diameter of the laser beam is given as 1.0 mm, so the radius would be 0.5 mm or \(0.5 * 10^{-3}\) m. We can calculate the area using the formula for the area of the circle, \( A = \pi r^{2}\). Substituting the radius value, we get: \( A = \pi ( 0.5 * 10^{-3} )^{2} = 7.85 * 10^{-7} \) square meters. Now, using the formula of intensity, \( I = P / A \), we can calculate the intensity of the laser beam at the laser: \( I = 5.0 * 10^{-3} / 7.85 * 10^{-7} = 6.37 * 10^{3} \) W/m².
02

Compute the intensity at the wall.

The spot on the wall is larger in diameter than the beam at the laser, which means that the beam's power is being spread over a larger area. The beam diameter at the wall is given as 3.6 cm or \(0.036\) m, so the radius is \(0.018 \) m. We can calculate the area using the formula for the area of a circle, \( A = \pi r^{2} \). So, \( A = \pi (0.018)^{2} = 0.00102 \) square meters. Now, using the formula of intensity, we can calculate the intensity of the laser beam at the wall: \( I = P / A \), thus, \( I = 5.0 * 10^{-3} / 0.00102 = 4.90 \) W/m².

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