A 20.0 W laser emits a bean of light 9 mm diameter. It is aimed at the moon by t

Question

A 20.0 W laser emits a bean of light 9 mm diameter. It is aimed at the moon by the time reaches the moon, it's has spread out to a diameter of 85 Km. what is the intensity of the light as it first leaves the laser? What is its inensity where it hits the moon? Assume there is no absorption and intensity units of power per units of area. A 20.0 W laser emits a bean of light 9 mm diameter. It is aimed at the moon by the time reaches the moon, it's has spread out to a diameter of 85 Km. what is the intensity of the light as it first leaves the laser? What is its inensity where it hits the moon? Assume there is no absorption and intensity units of power per units of area. A 20.0 W laser emits a bean of light 9 mm diameter. It is aimed at the moon by the time reaches the moon, it's has spread out to a diameter of 85 Km. what is the intensity of the light as it first leaves the laser? What is its inensity where it hits the moon? Assume there is no absorption and intensity units of power per units of area.

Explanation / Answer

Power P = 20 W

Diameter of beam on earth d = 9mm

radius: r = 9/2 = 4.5 mm

Intensity I = P / A = P / (4*pi*r2 ) = 20 /(4*3.14*4.52*10-6) = 0.0786 * 106 = 7.86 * 104 W/m2

Diameter of beam on moon d' = 85 km

radius r' = 85 /2 = 42.5 km

There was no absorption of beam so power remains same.

Thus Intensity on moon:

I' = P /A' = P / 4*pi*r'2 = 20 / (4*3.14*42.52 * 106) = 8.82 * 10-10 W/m2

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