Dept. of Physics & Astronomy

Office: Watanabe Hall, Rm. 426, (808) 956-2922

E-mail: nassir @ hawaii.edu

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PHYS 152 Past Exams & Answers*

**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)