Radiation of intensity 1.00 kW/m² is incident on an area perpendicular to the incoming light.The area is now rotated 20$^{\circ}$about an axis perpendicular to the incoming light.What is the resulting power per unit area now reaching the surface? A) 1.00 kW/m² B) 0.940 kW/m² C) 0.880 kW/m² D) 0.800 kW/m² E) 0.340 kW/m²

Question

Quizplus · Accepted Answer

When the area is rotated, the amount of radiation reaching the surface will depend on the angle between the normal to the surface and the direction of the incoming radiation. This angle is given by the cosine of the angle of rotation, which is cos(20°) ≈ 0.940.
Thus, the resulting power per unit area will be:
P' = P*cos(20°) = 1.00 kW/m² * 0.940 = 0.940 kW/m²
Therefore, the best choice is B.

Radiation of Intensity 1 ∘^{\circ}∘ About an Axis Perpendicular to the Incoming Light

Radiation of Intensity 1 $^{\circ}$ About an Axis Perpendicular to the Incoming Light