University of Hawaii, Manoa
Dept. of Physics & Astronomy

Instructor: Michael Nassir
Office: Watanabe Hall, Rm. 426, (808) 956-2922
E-mail: nassir @

Return to PHYS 152 Past Exams & Answers

ANSWERS to FINAL EXAM from Fall 2005

Part A

1a. 28 nA
1b. 5.1 x 10^2 keV (or 510 keV)
1c. 3.3 x 10^-25 kg
1d. 0.96 C/m^2
2a. C  (first combine 6-ohm and 8-ohm resistors in parallel, THEN combine with 12-ohm resistor in series)
2b. A  (must solve either for current through 12-ohm resistor (1.62 A), or for voltage drop across 12-ohm resistor (19.4 V), then use P = I*V = I^2*R = V^2/R)
3a. A  (C = Q/V)
3b. B  (U = (1/2)C*V^2 = (1/2)Q^2/C = (1/2)Q*V)
4a. A  (as B-field diminishes, change in flux is out-of-page, so emf is induced clockwise (Mike's Left-Hand Rule) )
4b. A  (again, as ring leaves B-field, B-field diminishes, change in flux is out-of-page, so emf is induced clockwise (Mike's Left-Hand Rule) )
5a. D  (1/s + 1/s' = 1/f, where f=0.210m and s'=12.0m)
5b. B  (real images are always inverted from the object's orientation, for one-lens systems)
6. D  (angular positions of, say, m=0 and m=1 maxima are found by: theta = arcsin(m * lambda / d) , and linear positions are found by: x = L * tan(theta) )

Part B

7a. C  (Wien's Law: peak wavelength = 2.898^10_3m*K / T)
7b. B
8. E  (K_max = (h*c/lambda) - W_0)
9. D
10. E
11. A
12. F   
13a. B  
13b. D  (delta E = (delta m) * c^2)
14. E
15. D  (lambda = (ln 2)/half-life;  N = N_0 * exp(-lamdba*t); from this we see that 0.785 of original sample remains undecayed, so 0.215 has decayed away)
16. B
17. B

Part C

1a. draw a few straight, evenly-spaced, horizontal E-field lines starting on right-hand plate and ending on left-hand plate
1b. draw three straight, evenly-spaced, vertical lines between the two plates: label 20V line 1/4 of way between plates, 40V line 1/2 way, and 60V line 3/4 of way between plates
1c. 2.6 x 10^-16 N, to the right (E = V/d in a uniform E-field = 1600 V/m; then F = q*E; since electron is negative, F is in opposite direction of E)
1d. A, B, C, D, F
1e. F

2a. 9.0 x 10^-5 T (or 0.90 G) into the page  (B = (mu_0 * I_1)/(2 * pi* r) )
2b. A  (B-field vectors add in same direction (into the page) )
2c. F  (B-field vectors add in opposite directions, canceling out each other and summing to zero)
2d. A  (magnetic force right-hand rule)
2e. 6.35 x 10^-19 C = 4.00*e  (F = q * v * B)
2f. A  (B-field pushes particle to the left, so we need an E-field that will exert a force on particle to the right)

Part D

3a. incoming ray passes through left-hand side of prism with NO bending -- continues in straight, horizontal line; ray bends downward toward lower-right (below horizontal) when exiting right-hand side of prism
3b. 31.9deg below horizontal  (ray is incident on RH side at 37.0deg; ray leaves RH side at 68.9deg below the normal; the normal is 37.0deg above horizontal, so emerging ray is 31.9deg below horizontal)
3c. B  (ray is totally internally reflected if critical angle is <= 37.0deg, which is true when n_glass > 1.66)
3d.i. B
3d.ii. B
3d.iii. C
3e. B (as the index of prism and index of surrounding medium become closer in value, the amount of refractive bending of ray becomes less)

4a. 6.04 x 10^14 Hz or 604 THz  (E = h*f, and E=2.50eV)
4b. 1240 nm  (for -1.00eV to -2.00eV transition)
and 355 nm  (for -1.00eV to -4.50eV transition)
4c. (a) transition upward from -4.50eV to -2.00eV; (b) transitions downward from -1.00eV to -2.00eV and from -1.00eV to -4.50eV
4d. part (a): visible (496nm);  part (b): infrared (1240nm) and ultraviolet (355nm)
4e. B  (electron must jump from -4.50eV level to 0.00eV level or higher)