Physics midterm question: same transmitted fraction of unpolarized/polarized light

Physics 205B Midterm 1, spring semester 2019
Cuesta College, San Luis Obispo, CA

In case (a) unpolarized light is incident on a 45° diagonal polarizer, and a certain fraction of incident light is transmitted through it. For case (b) horizontal polarized light is incident on a polarizer with a transmission axis that can be rotated to any θ angle. Discuss what the θ angle should be in case (b) such that the same fraction of incident light is transmitted through it as in case (a). Explain your reasoning using the properties of light and polarization.

Solution and grading rubric:

• p:
  Discusses/demonstrates that:
  1. in case (a), the fraction of unpolarized light that passes through a polarizer (regardless of its transmission angle) is (1/2); and
  2. for case (b), the fraction of polarized light that passes through a polarizer is given by Malus' law: cos²(90° − θ), where the angle of interest is between the horizontally polarization of the incident light and the transmission angle of the polarizer; and
  3. for the fraction transmitted in case (b) to equal the fraction transmitted in case (a), sets cos²(90° − θ) = (1/2), and thus θ = 45°.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. At least understands what happens in case (a), and has some systematic approach to match this fraction for case (b).
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at applying the properties of light, polarizers, and polarization.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. No clear attempt at applying the properties of light, polarizers, and polarization.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 30882, 30883
Exam code: midterm01Ft6G
p: 16 students
r: 2 students
t: 14 students
v: 10 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 7843):