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The intensity of sunlight hitting the earth is about 1300Wm2 . If sunlight strikes a perfect absorber, what pressure does it exert? How about a perfect reflector? What fraction of atmospheric pressure does this amount to?

Short Answer

Expert verified

The pressure in the case of a perfect absorber and reflector is 4.3×10-6N/m2and8.6×10-6N/m2 respectively, and the fraction of atmospheric pressure for a perfect absorber and the perfect reflector is 4.17×10-11and 8.3×10-11respectively.

Step by step solution

01

Expression for the pressure in case of a perfect absorber and a perfect reflector:

Write the expression for the pressure in the case of a perfect absorber.

P=lC ….. (1)

Here, Iis the intensity, and c is the speed of light.

Write the expression for the pressure in the case of a perfect reflector.

P=2lC ….. (2)

02

Determine the pressure in case of a perfect absorber and a perfect reflector:

Substitute the values in equation (1).

P=1300W/m23×108m/sP=4.3×10-6N/m2

Substitute the values in equation (2).

P'=21300W/m23×108m/sP'=8.6×10-6N/m2

03

Determine the fraction of atmospheric pressure:

The atmospheric pressure is equal to 1.03×105N/m2
.Calculate the fraction of atmospheric pressure for a perfect absorber.

Fabsorber=4.3×10-6N/m21.03×105N/m2Fabsorber=4.17×10-11

Calculate the fraction of atmospheric pressure for a perfect reflector.

Fabsorber=8.6×10-6N/m21.03×105N/m2Fabsorber=8.3×10-11

Therefore, the pressure in the case of a perfect absorber and reflector isrole="math" localid="1657366852692" 4.3×10-6N/m2 and8.6×10-6N/m2 respectively, and the fraction of atmospheric pressure for a perfect absorber and the perfect reflector is4.17×10-11 and 8.3×10-11respectively.

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