Unpolarized light passes through two Polaroid sheets.

Unpolarized light passes through two Polaroid sheets. The transmission axis of the analyzer makes an angle of 42.8° with the axis of the polarizer. (a) What fraction of the original unpolarized light is transmitted through the analyzer? (Enter your answer to at least three decimal places.) (b) What fraction of the original light is absorbed by the analyze

The Correct Answer and Explanation is:

This is a classic problem of light passing through polarizers. The fraction of light transmitted through the analyzer can be found using Malus’ law, which states that: I=I0cos⁡2θI = I_0 \cos^2 \thetaI=I0​cos2θ

Where:

• III is the intensity of light after passing through the analyzer.
• I0I_0I0​ is the intensity of the light before passing through the analyzer.
• θ\thetaθ is the angle between the transmission axis of the polarizer and analyzer.

Part (a) Fraction of light transmitted through the analyzer

1. Step 1: Use Malus’ Law
   The light is initially unpolarized, so after passing through the first Polaroid (the polarizer), it becomes polarized. The transmission axis of the analyzer makes an angle of 42.8∘42.8^\circ42.8∘ with the axis of the polarizer. We use Malus’ law to find the fraction of light transmitted.

I=I0cos⁡2(42.8∘)I = I_0 \cos^2(42.8^\circ)I=I0​cos2(42.8∘)

2. Step 2: Calculate cos⁡(42.8∘)\cos(42.8^\circ)cos(42.8∘)
   Using a calculator:

cos⁡(42.8∘)≈0.743\cos(42.8^\circ) \approx 0.743cos(42.8∘)≈0.743

So, I=I0×(0.743)2=I0×0.552I = I_0 \times (0.743)^2 = I_0 \times 0.552I=I0​×(0.743)2=I0​×0.552

Thus, the fraction of the original unpolarized light transmitted through the analyzer is approximately: 0.552\boxed{0.552}0.552​

Part (b) Fraction of light absorbed by the analyzer

The light absorbed by the analyzer is simply the fraction that does not pass through. Therefore, the fraction of light absorbed is: Absorbed fraction=1−Transmitted fraction=1−0.552=0.448\text{Absorbed fraction} = 1 – \text{Transmitted fraction} = 1 – 0.552 = 0.448Absorbed fraction=1−Transmitted fraction=1−0.552=0.448

So, the fraction of light absorbed by the analyzer is approximately: 0.448\boxed{0.448}0.448​

Summary:

• The fraction of the original unpolarized light transmitted through the analyzer is 0.5520.5520.552.
• The fraction of the original light absorbed by the analyzer is 0.4480.4480.448.