Download Sample Test Ch 23-28

Grade: _______
Phys. 245…Summer 2008…Test
Chapters 23-28 …Allotted time: 1 hr 30 m
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Guidelines
1) All work is to be done on the pages provided here. If needed, problem solutions can be continued on the
back of the page, being sure that your solution is easily traceable by the grader. If your problem solution is
not easily traceable, the problem’s point value will be sacrificed.
2) Final answers are to be clearly entered in the spaces provided, with boxes around answers to the
quantitative free-form problems.
3) This is a closed-book, no-notes-of-any-kind test. Only the physical constants typically required in the
problem solutions are provided in the table below, followed by a list of principle equations for the chapters
covered in this test
4) Use of calculators is (of course) allowed, but they cannot be programmed or be loaded in anyway with
formulae or solution guidelines relevant to this test and this course.
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“Summary” Equations: Ch. 19 (DC Circuits)
(19-3) Req = R1 + R2 + R3
(19-4) (1/ Req ) = (1/ R1) + (1/ R2 ) + (1/ R3 )
(19-6) (1/Ceq ) = (1/C1) + (1/C2) + (1/C3) (19-7) τ = RC
(19-5) Ceq = C1 = C2 + C3
“Summary” Equations: Ch. 20 (Magnetism)
(20-1) F = IlB sin θ
(20-10) τ = NIAB sin θ
(20-2) F = qvB sin θ
Ex.20-3) qEv = mv2/r
(20-6) B = (µ0/2π)I/r
(20-8) B = µ0 NI/l
“Summary” Equations: Ch. 21 (Electromagnetic Induction and Faraday’s Law)
(21-1) ΦB = B┴ A = BA cos θ
(21-7) Is/Ip= Np/Ns
(21-2b) E = -N ∆ΦB/∆t
(21-3) E = Blv
(21-6) Vs/Vp = Ns/Np
“Summary” Equations: Ch. 22 (Electromagnetic Waves)
(22-3) c = 1/(ε0µ0)1/2
(22-4) c = λf
“Summary” Equations: Ch. 23 (Light – Geometric Optics)
(23-1) f = r/2
(23-5) n1 sin θ1 = n2 sin θ2
(23-8) m = hi /ho = - di /do
(23-2) (1/do) + (1/di) = 1/f
(23-6) sin θc = n2 / n1
(23-3) m = hi /ho = - di /do
(23-7) (1/do) + (1/di) = 1/f
(23-4) n = c/v
“Summary” Equations: Ch. 24 (Wave Nature of Light)
(24-1) λn = λ/n
(24-3a) sin θ = λ/d
(24-2a) sin θ = m λ/d
(24-5) I = Io cos2 θ
(24-2b) sin θ = (m + ½)λ/d
(24-6b) tan θp = n
“Summary” Equations: Ch. 25 (Optical Instruments)
(25-2a) M = N/f
(25-3) M = -fo/fe
(25-6b) M ≈ Nl/ fofe
“Summary” Equations: Ch. 26 (Special Theory of Relativity)
(26-1) ∆t = ∆to/(1 – (v/c)2)1/2 = γto
(26-3) L = Lo(1 – (v/c)2)1/2 = Lo/γ
(26-2) γ = 1/(1 – (v/c)2)1/2
2 1/2
2 1/2
(26-4) p = γmov = mov/(1 – (v/c) )
(25-5) mrel = γmo = mo/(1 – (v/c) )
(26-8) Eo = moc2
(26-6) KE = moc2/(1 – (v/c)2)1/2 - moc2 = (γ - 1) moc2
(26-7) E = KE + moc2 = γ moc2
(26-10) E2 = p2c2 + mo2c4
“Summary” Equations: Ch. 27 (Early Quantum Theory and Models of the Atom)
(27-1) E = nhf
(27-11) L = nh/2π
(27-4) E = hf
(27-6) p = E/c = hf/c = h/ λ
(27-14) rn = (n2/Z)r1 = (n2/Z)(0.529x10-10m) n = 1, 2, 3,…
(27-8) λ = h/p
(27-15b) En = - (13.6eV)Z2/n2
“Summary” Equations: Ch. 28 (Quantum Mechanics of Atoms)
(28-1) (∆px)( ∆x) ≥ h/2 π
(28-2) (∆E)( ∆t) ≥ h/2 π
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Part 1 – Multiple Choice
15 Questions…2 Points Each…Total Points = 30
1) Is it possible to see a virtual image?
A) No, since the rays that seem to emanate from a virtual image do not in fact emanate from the image.
B) No, since virtual images do not really exist.
C) Yes, the rays that appear to emanate from a virtual image can be focused on the retina just like those from an
illuminated object.
D) Yes, since almost everything we see is virtual because most things do not themselves give off light, but only
reflect light coming from some other source.
E) Yes, but only indirectly in the sense that if the virtual image is formed on a sheet of photographic film, one
could later look at the picture formed.
Answer: C
2) A light ray, traveling parallel to a concave mirror's axis, strikes the mirror's surface near its midpoint. After
reflection, this ray
A) again travels parallel to the mirror's axis.
B) travels at right angles to the mirror's axis.
C) passes through the mirror's center of curvature.
D) passes through the mirror's focal point.
Answer: D
3) A convex lens has focal length f. An object is placed at 2f on the axis. The image formed is located
A) at 2f.
B) between f and 2f.
C) at f.
D) between the lens and f.
Answer: A
4) Two light sources are said to be coherent if they
A) are of the same frequency.
B) are of the same frequency, and maintain a constant phase difference.
C) are of the same amplitude, and maintain a constant phase difference.
D) are of the same frequency and amplitude.
Answer: B
5) We have seen that two monochromatic light waves can interfere constructively or destructively, depending
on their phase difference. One consequence of this phenomenon is
A) the colors you see when white light is reflected from a soap bubble.
B) the appearance of a mirage in the desert.
C) a rainbow.
D) the way in which Polaroid sunglasses work.
E) the formation of an image by a converging lens, such as the lens in your eye.
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Answer: A
6) The principal refraction of light by the eye occurs at the
A) cornea.
B) lens.
C) retina.
D) iris.
Answer: A
7) What type of lens is a magnifying glass?
A) converging
B) diverging
C) spherical
D) cylindrical
Answer: A
8) You are riding in a spaceship that has no windows, radios, or other means for you to observe or measure
what is outside. You wish to determine if the ship is stopped or moving at constant velocity. What should you
do?
A) You can determine if the ship is moving by determining the apparent velocity of light.
B) You can determine if the ship is moving by checking your precision time piece. If it's running slow, the ship
is moving.
C) You can determine if the ship is moving either by determining the apparent velocity of light or by checking
your precision time piece. If it's running slow, the ship is moving.
D) You should give up because you have taken on an impossible task.
Answer: D
9) One of Einstein's postulates in formulating the special theory of relativity was that the laws of physics are the
same in reference frames that
A) accelerate.
B) move at constant velocity with respect to an inertial frame.
C) oscillate.
D) are stationary, but not in moving frames.
Answer: B
10) If the wavelength of a photon is halved, by what factor does its energy change?
A) 4
B) 2
C) 1/4
D) 1/2
Answer: B
11) When a photon is scattered from an electron, there will be an increase in its
A) energy.
B) frequency.
C) wavelength.
D) momentum.
Answer: C
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12) In the ground state, the quantum numbers (n, l, ml, ms) for hydrogen are, respectively,
A) 1, 1, 1, 1.
B) 1, 0, 0, 0.
C) 1, 0, 0, ±1/2.
D) 1, 1, 1, ±1/2.
Answer: C
13) In terms of an atom's electron configuration, the letters K, L, M, and N refer to
A) different shells with n equal to 1, 2, 3, or 4 respectively.
B) different sub shells with l equal to 1, 2, 3, or 4 respectively.
C) the four possible levels for the magnetic quantum number.
D) the four possible quantum numbers.
Answer: A
14) The hydrogen nucleus consists of
A) a single proton.
B) a single neutron.
C) one proton and one neutron.
D) one proton and two neutrons.
Answer: A
15) The reason the position of a particle cannot be specified with infinite precision is the
A) exclusion principle.
B) uncertainty principle.
C) photoelectric effect.
D) principle of relativity.
Answer: B
Part 2 – Quantitative Free-Form Problems
70 Total Points…Each Problem Value as Noted
PUT A BOX AROUND ALL ANSWERS
16) (8 pts) A person whose eyes are 1.68 m above the floor stands 2.20 m in front of a vertical plane mirror
whose bottom edge is 43 cm above the floor, Fig. 23–48. What is the horizontal distance x to the base of
the wall supporting the mirror of the nearest point on the floor that can be seen reflected in the mirror?
Solution:
The angle of incidence is the angle of reflection.
Thus we have
( H − h) h
tan θ =
= ;
L
x
(1.68m − 0.43m ) (0.43m)
=
,
x
( 2.20 m )
which gives x = 0.76 m = 76cm.
Mirror
H
θ
θ
h
x
L
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17) (4 pts) The speed of light in ice is 2.29 × 10 8 m s . What is the index of refraction of ice?
Solution:
We find the index of refraction from
c
v= ;
n
3.00 × 108 m s )
(
8
2.29 × 10 m s =
, which gives n = 1.31.
n
18) (10 pts) Monochromatic light falls on two very narrow slits 0.048 mm apart. Successive fringes on a screen
5.00 m away are 6.5 cm apart near the center of the pattern. Determine the wavelength and frequency of the
light.
Solution: For constructive interference, the path difference is a multiple of the wavelength:
d sin θ = mλ , m = 0, 1, 2, 3, ... .
We find the location on the screen from
y = L tan θ .
For small angles, we have
sin θ ≈ tan θ , which gives
⎛ mλ ⎞ mLλ
y = L⎜
⎟=
⎝ d ⎠
.
d
For adjacent fringes, ∆m = 1, so we have
Lλ ∆m
∆y =
;
d
0.065m =
( 5.00 m ) λ (1) , which gives λ =
−3
( 0.048 × 10
m)
6.2 × 10−7 m.
The frequency is
( 3.00 × 10 m s ) =
f = =
λ ( 6.24 × 10 m )
8
c
−7
4.8 × 1014 Hz.
19) (12 pts) A certain star is 10.6 light-years away. How long would it take a spacecraft traveling 0.960c to
reach that star from Earth, as measured by observers: (a) on Earth, (b) on the spacecraft? (c) What is the
distance traveled according to observers on the spacecraft? (d) What will the spacecraft occupants compute
their speed to be from the results of (b) and (c)?
Solution: (a) To an observer on Earth, 10.6 ly is the rest length, so the time will be
tEarth =
L0 (10.6ly )
=
= 11.0 yr.
v
0.960c
(b) We find the dilated time on the spacecraft from
∆t =
∆t 0
⎡ ⎛ v ⎞⎤
⎢1 − ⎜ 2 ⎟ ⎥
⎣ ⎝ c ⎠⎦
2
1
2
;
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∆t0
11.0 yr =
⎡1 − ( 0.960 ) ⎤
⎣
⎦
2
1
2
, which gives ∆t0 = 3.09 yr.
(c) To the spacecraft observer, the distance to the star is contracted:
1
1
⎡ ⎛ v ⎞2 ⎤ 2
2 2
⎡
⎤
= 2.97 ly.
L = L0 ⎢1 − ⎜ ⎟ ⎥ = (10.6ly ) 1 − ( 0.960 )
⎣
⎦
⎢⎣ ⎝ c ⎠ ⎥⎦
(d) To the spacecraft observer, the speed of the spacecraft is
v=
L ( 2.97 ly )
=
= 0.960c, as expected.
∆t
3.09 yr
20) (4 pts) The total annual energy consumption in the United States is about 8 × 1019 J. How much mass would
have to be converted to energy to fuel this need?
Solution: We find the necessary mass conversion from
8 × 1019 J )
(
∆E
∆m = 2 =
= 9 × 102 kg.
2
8
c
( 3.00 × 10 m s )
21) (12 pts) Estimate the peak wavelength for radiation from (a) ice at 0°C, (b) a floodlamp at 3500 K, (c)
helium at 4 K, (d) for the universe at T = 2.725 K, assuming blackbody emission. In what region of the EM
spectrum is each?
Solution: (a) We find the peak wavelength from
λP =
( 2.90 × 10
−3
miK )
T
=
( 2.90 × 10
−3
mi K )
( 273K )
= 1.06 × 10−5 m = 10.6 µ m.
This wavelength is in the near infrared.
(b) We find the peak wavelength from
λP =
( 2.90 × 10
−3
miK )
T
=
( 2.90 × 10
−3
mi K )
( 3500 K )
= 8.29 × 10−7 m = 829 nm.
This wavelength is in the infrared.
(c) We find the peak wavelength from
λP =
( 2.90 × 10
−3
miK )
T
=
( 2.90 × 10
−3
(4K)
mi K )
= 7.25 × 10−4 m = 0.73mm.
This wavelength is in the microwave region.
(d) We find the peak wavelength from
λP =
( 2.90 × 10
T
−3
mi K )
=
( 2.90 × 10
−3
miK )
( 2.725K )
= 1.06 × 10−2 m = 1.06 cm.
This wavelength is in the microwave region.
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22) (6 pts) The human eye can respond to as little as 10 −18 J of light energy. For a wavelength at the peak
of visual sensitivity, 550 nm, how many photons lead to an observable flash?
Solution: The energy of a single photon is
E = hf =
hc
λ
( 6.63 × 10 Jis )( 3.00 × 10
=
( 550 × 10 m )
−34
8
m s)
−9
= 3.62 × 10−19 J.
Therefore, the number of photons required for an observable flash is calculated as
10−18 J
= 2.8.
3.62 × 10−19 J
Rounding up, we find that 3 photons are required.
23) (8 pts) A hydrogen atom is in the 6s state. Determine (a) the principal quantum number, (b) the energy of
the state, (c) the orbital angular momentum and its quantum number l, and (d) the possible values for the
magnetic quantum number.
Solution: (a) The principal quantum number is n = 6.
(b) The energy of the state is
E6 = −
(13.6eV ) = − (13.6eV ) =
n2
62
− 0.378eV.
(c) The “s” subshell has = 0. The magnitude of the angular momentum
depends on only:
L = ˜ ⎡⎣
(
1
1
+ 1) ⎤⎦ 2 = (1.055 × 10−34 Jis ) ⎡⎣( 0 )( 0 + 1) ⎤⎦ 2 = 0.
(d) For each
the value of m can range from − to + : m = 0.
⎧
⎫
⎪
⎪⎪
1
⎪
= (1.67 × 10−27 kg )( 2.60 × 108 m s ) ⎨
= 8.70 × 10−19 kgi m
1 ⎬
⎪ ⎡1 − 0.867 2 ⎤ 2 ⎪
(
) ⎦ ⎭⎪
⎩⎪ ⎣
⎛ 1.63 × 103 MeV ⎞
s⎜
⎟.
c
⎝
⎠
24) (6 Pts) A certain lens focuses light from an object 2.75 m away as an image 48.3 cm on the other side of the
lens. What type of lens is it and what is its focal length? Is the image real or virtual?
Solution: To form a real image from a real object requires a converging lens.
We find the focal length of the lens from
⎛ 1 ⎞ ⎛1⎞ 1
⎜ ⎟ +⎜ ⎟= ;
⎝ d o ⎠ ⎝ di ⎠ f
⎛ 1 ⎞ ⎛ 1 ⎞ 1
⎜
⎟+⎜
⎟ = , which gives f = + 41.1cm.
⎝ 275cm ⎠ ⎝ 48.3cm ⎠ f
Because di > 0, the image is real.
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