20160215

Physics quiz question: possible total internal reflection?

Physics 205B Quiz 1, spring semester 2016
Cuesta College, San Luis Obispo, CA

A beam of light strikes the interface between room temperature water (index of refraction[*]1.333) and diamond (index of refraction 2.418), with an incident angle in water of 33.0°. (Drawing is not to scale.) If the incident angle of the beam of light in water is increased slightly from 33.0° to 33.5°, the incident beam would be:
(A) reflected back into water.
(B) transmitted into diamond.
(C) (Both of the above choices.)
(D) (Not enough information is given.)

[*] physics.info/refraction/.

Correct answer (highlight to unhide): (C)

Snell's law relates the indices of refraction with the incident and transmitted angles:

nwater·sin(θwater) = ndiamond·sin(θ diamond ),

where the given (or assumed to be known) quantities, unknown quantities, and quantities to be explicitly solved for are denoted.

Directly solving for the transmitted angle (if any) in diamond:

sin(θdiamond) = (nwater/ndiamond)·sin(θwater),

θdiamond = sin-1(nwater/ndiamond)·sin(θdiamond)) = sin–1((1.333/2.418)·sin(33.5°)) = 17.7144284003°,

or to three significant figures, the transmitted angle in diamond is 17.7°. Thus the light will be both reflected back into water and transmitted into diamond.

(The critical angle for light that is instead originating in diamond, incident on a diamond-water interface is given by:

θc = sin–1(nwater/ndiamond) = sin–1(1.333/2.418) = 33.46°.

Since the incident angle of 33.5° is greater than the critical angle, then the light that is originating in diamond will experience total internal reflection back down into the diamond, with none of it transmitted out into water).

Student responses
Section 70856
Exam code: quiz01sN0w
(A) : 6 students
(B) : 6 students
(C) : 30 students
(D) : 1 student

Success level: 70%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.43

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