Unpolarized light with an average intensity of 845 W/m² enters a polarizer with a vertical transmission axis. The transmitted light then enters a second polarizer. The light that exits the second polarizer is found to have an average intensity of 225 W/m². What is the orientation angle of the second polarizer relative to the first one? A)54.7° B)22.6° C)43.1° D)77.3° E)zero degrees

Question

Quizplus · Accepted Answer

When unpolarized light passes through a polarizer with transmission axis at an angle θ to the vertical, the intensity of the transmitted light is given by:

I = I0 cos²(θ).

In this problem, the first polarizer has a vertical transmission axis (θ = 0) and the intensity of the light after passing through it is:

I1 = I0 cos²(0) = I0.

The transmitted light then passes through a second polarizer with transmission axis at an angle θ2 to the vertical. The intensity of the transmitted light is again given by:

I2 = I1 cos²(θ2).

We are given that I1 = 225 W/m² and I0 = 845 W/m², so:

225 = 845 cos²(θ2)

cos²(θ2) = 225/845 = 0.266

Taking the square root of both sides and using the fact that 0° ≤ θ2 ≤ 90°, we get:

cos(θ2) = 0.516

θ2 ≈ 43.1°

Therefore, the orientation angle of the second polarizer relative to the first one is 43.1°.

Unpolarized Light with an Average Intensity of 845 W/m2 Enters

Unpolarized Light with an Average Intensity of 845 W/m² Enters