Physics 112
Exam 3
Spring 2015
33. Blue/violet light has a frequency of 7.41 × 1014 Hz in vacuum. What is the
FREQUENCY (in Hz) of this light when it travels through glass with a refractive index
of 1.51?
B. 4.81 × 1014
A. zero
C. 7.41 × 1014
D. 1.01 × 1014
E. 400
There is no change in the frequency when light
passes from one transparent material to another
34. Light from a red laser (λ = 650 nm) illuminates a SINGLE slit. A diffraction pattern
is seen on the wall 12.0 m away. The central peak has a FULL WIDTH of 14.0 cm
What is the WIDTH of the slit?
(Full width is measured from 1st minimum to 1st minimum).
A. 0.111 mm
B. 1.67 mm
slit
12.0 m
C. 28.2 mm
D. 5.53 cm Minima in single slit diffraction pattern are given by,
E. 17.1 cm
= λ, where m = 1, 2, 3…
14.0 cm
wall
Rearranging for D and using the small angle limit gives,
λ (1)(650 × 10 )(12)
=
=
= 1.11 × 10 (0.07)
35. Light is incident (normally) on a transmission GRATING with 8900 lines/cm. FIRST
order lines are seen at angles: θ = 25.2o and 27.1o after the grating.
What WAVELENGTHS (in nm) are emitted by the light source?
A.
B.
C.
D.
E.
478
478
504
504
405
and
and
and
and
and
596
512
512
590
532
source
Maxima in a grating diffraction
pattern are given by,
= λ, where m = 1, 2, 3…
Rearranging for λ gives,λ =
= (
grating
θ
(note, small angle limit doesn’t apply
because the angles are too big),
)(!)
Substitute for the three values of θ to give, λ = 478, 512, 689 nm
36. What is the minimum thickness of an OIL film (n = 1.51) on WATER (n = 1.33) if
the film appears bright when light of wavelength 532 nm is incident on it?
(assume normal incidence)
A.
B.
C.
D.
E.
88 nm
100 nm
174 nm
266 nm
300 nm
Ray 1 experiences a phase shift of
π (≡ λ/2) on reflection, ray 2 does not
Ray 2 must see a path difference of
!
2" = ( + $)λ%& where λ%& = λ⁄%& , in
order to produce constructive interference
At the min thickness, m = 0.
()$
So, " = ()(!.(!) = 88.1
1 π
2 0
oil
water
Physics 112
Exam 3
Spring 2015
37. HORIZONTALLY POLARIZED light is shone on two polarizers. The transmission
axes of the sheets are aligned at 60o and 20o to the horizontal axis, as shown.
What PERCENTAGE of the incident light is TRANSMITTED through both polarizers?
A. 65 %
B. 54 %
C. 23 %
D. 15 %
E. 0
60ο
I 2 = cos 2 (60)cos 2 (40) = 0.147 I 0
20ο
Horizontally
polarized light
38. When a CONVERGING lens is used as a SIMPLE MAGNIFIER you see an
A. UPRIGHT, REAL, magnified image
B. INVERTED, REAL, magnified image
C. UPRIGHT, VIRTUAL, magnified image
D. INVERTED, VIRTUAL, magnified image
E. REAL, VIRTUAL, magnified image
FΟ
Ι
39. Emily has a FAR POINT of 25 cm and a NEAR POINT of 10 cm.
What TYPE and POWER eyeglass lenses should be prescribed to allow her to see distant
objects clearly?
(assume the lenses are 2.0 cm from Emily’s eyes)
A.
B.
C.
D.
E.
!
+,
6.0 Diopters (Converging)
4.7 Diopters (Converging)
1.7 Diopters (Converging)
− 3.0 Diopters (Diverging)
− 4.3 Diopters (Diverging)
!
!
+ + −= /
-
!
25 cm
!
+ .$) = −4.3
0
→
40. A microscope consists of an objective lens (focal length = 1.0 cm) and an eyepiece
(focal length = 2.8 cm). The object is located 1.2 cm from the objective. The distance
between the lenses is adjusted for a relaxed normal eye, i.e. the final image is at infinity.
What is the DISTANCE between the LENSES after the adjustment?
A. 1.2 cm
l
B. 2.4 cm C. 4.0 cm D. 8.8 cm E. 15 cm
objective
For the objective,
!
!
!
+ + = !.
!.$
-
→
3 = 6.04
Fo
For a relaxed eye the first image is at Fe so,
5 = 6.0 + 2.8 = 8.84
6.0 cm
Final image at ∞
eyepiece
2.8 cm
Fe
Physics 112
Exam 3
Spring 2015
41. A muon is an elementary particle with a rest mass of 105.7 MeV/c2.
If a particular muon has a kinetic energy of 160 MeV, what is γ for this muon?
A. 1.00
B. 1.60
67 = (8 − 1)% 4 $
C. 2.51
D. 15.4
160 = (8 − 1)105.7
→
→
E. 22.6
8 = 2.51
42. A suitcase is 1.0 m long and 0.5 m high when it is measured at rest. Imagine YOU
take the suitcase on a high speed train that moves at 0.6c. YOU measure the dimensions
of the suitcase again on the train (suitcase handle always on top, as shown).
Which statement is TRUE?
YOU measure the suitcase to be,
A. MORE than 1.0 m long, but still 0.5 m high.
B. 1.0 m long and 0.5 m high.
C. LESS than 1.0 m long, but still 0.5 m high.
D. LESS than 1.0 m long and LESS than 0.5 m high
E. It is impossible to say from the information given.
The suitcase is at rest relative to you so you measure its proper length
43. An electron has momentum of 0.846 MeV/c, and a rest mass of 0.511 MeV/c2.
What is the KINETIC ENERGY of this electron?
A. 1.36 MeV
B. 0.924 MeV
C. 0.658 MeV
D. 0.533 MeV
E. 0.477 MeV
7 = 9:$ 4 $ + % $ 4 = √0.846$ + 0.511$ = 0.9884=>?
7 = 67 + % 4 $ →
67 = 0.9884 − 0.511 = 0.477=>?
44. What is the de Broglie WAVELENGTH for a pollen grain (mass = 25 x 10-12 kg)
blowing in the wind at 5.5 m/s?
A. 5.9 × 10-36 m
B. 6.6 × 10-28 m
C. 5.3 × 10-24 m
D. 4.8 × 10-24 m
E. 9.5 × 10-6 m
@
B.B)×! CDE F
λ = A = ($(×!
CGH IJ)((.(K)
L
= 4.8 × 10$ Physics 112
Exam 3
Spring 2015
45. UV photons (λ = 184 nm) are incident on a metal surface in a vacuum and electrons
with a maximum kinetic energy of 1.64 eV are emitted from the surface.
What is the WORK FUNCTION (in eV) of the metal?
A. 5.10
@M
λ
B. 4.65
= 67NO + P%
→
C. 4.08
!.$×! D QR
!
D. 3.63
= 1.64>? + P%
E. 2.30
→ P% = 6.74 − 1.64 = 5.10>?
46. The following is a list of electronic transitions in the Bohr model of a hydrogen atom.
The first number is the initial value of n and the second is the final value of n.
In which transition does the electron GAIN the most energy?
(you should be able to answer this without a calculation)
A. 8 → 1
B. 2 → 5
C. 4 → 7
D. 4 → 2
E. 16 → 6
4
3
If the electron gains energy, n must increase,
2
so only B and C are possible
1
From the diagram, a change from n = 2 to 5
involves a bigger jump than 4 to 7
1 1
 1 1 
Calculation: ∆E = 13.6 eV  2 − 2  = 2.9 eV is bigger than ∆E = 13.6 eV  2 − 2  = 0.57 eV
2 5 
4 7 
47. How much TOTAL KINETIC ENERGY will an electron-positron PAIR have if they
are produced by a 2.452 MeV photon?
(the electron and positron each have a rest mass 0.511 MeV/c2).
A. 1.90 MeV
B. 0.500 MeV
C. 1.52 MeV
D. 0.892 MeV
E. 1.43 MeV
Energy conservation:
energy of photon = total KE + total rest mass energy of particles
7 = 67S%SN& + 2% 4 $
→
67S%SN& = 2.452=>? − 2(0.511) = 1.43=>?
48. This is version B of the exam. Shade in circle B on number 48.
for the pair