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 and for polycarbonate, are:

nair = c/vair,

npolycarbonate = c/vpolycarbonate,

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:

λair = vair/fair,

λpolycarbonate = vpolycarbonate/fpolycarbonate.

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

fpolycarbonate = fair,

fpolycarbonate = vair/λair,

fpolycarbonate = (c/nair)/λair,

fpolycarbonate = c/(nair·λair) = (3.00×108 m/s)/((1.000)·(405×10–9 m)),

fpolycarbonate = 7.40740740741×1014 s–1,

or to three significant figures, 7.41×1014 Hz.

(Response (A) is c/(λair·npolycarbonate); response (C) is (c·npolycarbonate)/λair; 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

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