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Shorty43 · May 7, 2012, 06:56 AM

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 a diameter of 85km. 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?

ebaines · May 10, 2012, 12:08 PM

Originally Posted by Shorty43

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 a diameter of 85km. 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?

The intensity of the light can vbe expressed in terms of watts per square meter. So divide the power of the laser, which is 10 watts, by the area over which that 10 watts is spread. Recall that the area of a circle of diameter D is $A=\frac 1 4 \pi D^2$ .

ebaines · May 10, 2012, 12:09 PM

Originally Posted by ebaines

The intensity of teh light can vbe expressed in terms of watts per square meter. So divide the power of the laser, which is 10 watts, by the area over which that 10 watts is spread. Recall that the area of a circle of diameter D is $A=\frac 1 4 \pi D^2$ .

. Be sure to express the area in square meters.