## 20150213

### Physics quiz question: frequency of Blu-ray™ laser light in polycarbonate

Physics 205B Quiz 1, spring semester 2015
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Comprehensive Problem 22.67(a)

A Blu-ray Disc™ player uses laser light (wavelength[*] 405 nm) in air (index of refraction of 1.000) that goes through a layer of polycarbonate (index of refraction[**] of 1.585). The frequency of this light in polycarbonate is:
(A) 4.67×1014 Hz.
(B) 7.41×1014 Hz.
(C) 1.17×1015 Hz.
(D) 1.22×1020 Hz.

[*] en.wikipedia.org/wiki/File:Comparison_CD_DVD_HDDVD_BD.svg.
[**] physics.info/refraction/.

Correct answer (highlight to unhide): (B)

The relations between the index of refraction and the speed of light, for air (medium 1) and for polycarbonate (medium 2), are:

n1 = c/v1,

n2 = c/v2,

where the given (or assumed to be known) quantities, unknown quantities, and quantities to be explicitly solved for are denoted. Also the relations between wavelength, speed, and frequency are:

λ1 = v1/f1,

λ2 = v2/f2.

However, the frequency of the light in air is the same as the frequency it has in polycarbonate, such that:

f2 = f1,

f2 = v1/λ1,

f2 = (c/n1)/λ1,

f2 = c/(n1·λ1) = (3.00×108 m/s)/((1.000)·(405×10–9 m)),

f2 = 7.40740740741×1014 s–1,

or to three significant figures, 7.41×1014 Hz.

(Response (A) is c/(λ1·n2); response (C) is (c·n2)/λ1; response (D) is c·(405×109 m).)

Sections 30882, 30883
Exam code: quiz01c0Co
(A) : 24 students
(B) : 21 students
(C) : 3 students
(D) : 0 students

Success level: 44%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.41