Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 23.11

(A) 22.1 degrees.
(B) 40.6 degrees.
(C) 41.7 degrees.
(D) 48.8 degrees.
Correct answer: (C)
From Snell's law, n_i*sin(theta_i) = n_t*sin(theta_t), where n_i = 1.000 (given), n_t = 1.33, and theta_t = 30.0 degrees. Then solving for the incident angle in air:
theta_i = Arcsin(n_t*sin(theta_t)/n_i) = 41.7 degrees.
Response (A) is Arcsin(sin(theta_t)/n_t); response (B) is Arcsin(sin(90 degrees - theta_t)/n_t); response (D) is Arcsin(n_i/n_t), which is the critical angle for water incident on air.
Student responses
Section 31988
(A) : 2 students
(B) : 0 students
(C) : 10 students
(D) : 1 student
Success level: 76%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.50
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