20100604

Physics final exam problem: anti-glare thin film

Physics 205B Final Exam, Spring Semester 2010
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problems 25.19, 25.21

[20 points.] Glass (n = 1.52) is coated with a thin film of magnesium fluoride (n = 1.38) of thickness 125 nm. Which visible (400-700 nm) wavelength(s) in air reflects with destructive interference? Show your work and explain your reasoning.

Solution and grading rubric:
  • p = 20/20:
    Correct. Recognizes (a) reflection off of top of MgF film causes 180 degree phase shift, as well as the reflection off of bottom of MgF film, such that the two reflected rays are in-of-phase with each other; (b) the path difference is 2*t = 125 nm; (c) for destructive interference of two in-of-phase reflected rays, delta(l) = (m + 1/2)*lambda = odd*lambda/2, where lambda is the wavelength in MgF; (d) wavelength in MgF is related to the wavelength in air by lambda = lambda_0/n_MgF, such that lambda_0 = 4*n*t/odd = 690 nm, 230 nm, ..., of which only 690 nm is in the visible light range.
  • r = 16/20:
    Nearly correct, but includes minor math errors. Omission of one or two of concepts (a)-(d) outlined above.
  • t = 12/20:
    Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. Demonstrates some understanding of at least one of concepts (a)-(d) outlined above.
  • v = 8/20:
    Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner.
  • x = 4/20:
    Implementation of ideas, but credit given for effort rather than merit.
  • y = 2/20:
    Irrelevant discussion/effectively blank.
  • z = 0/20:
    Blank.

Grading distribution:
Section 31988
p: 3 students
r: 2 students
t: 2 students
v: 5 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 1987):


A sample "r" response (from student 6686), solving for the wavelength in MgF, but not relating this to the wavelength value it would have in air:

No comments: