Dr. Huerta PHY 206 Sample Final Spring 2010

Dr. Huerta
PHY 206 Sample Final
Spring 2010
Signature:
Spring 2009, Section FJ, MWF 1:25 – 2:15 p. m.
Name:
1
Idnumber:
[ ] Dr. Brown 1O, 9:30 - 10:20 a.m.
[ ] Dr. Brown 1P, 11:00 - 11 :50a.m.
2
[ ] Dr. Zuo 3P, 11:00 - 11 :50a.m.
[ ] Mr. Sarisaman 1Q, 12:30 - 1:20 p.m.
3
[ ] Dr. Huerta 1R, 2:00 - 2:50 p.m.
[ ] Dr. Zuo 1S, 3:30 - 4:20 p.m.
4
Z
p = mv,
f
Impulse =
Fdt = ∆p = Fav ∆t.
5
i
Use g = 10
m
,
s2
dp
= −ρg,
dy
p(y) = p0 −ρgy,
ρwater = 1 × 103
kg
,
m3
Patm = 105 Pa,
ρ 1 A1 v 1 = ρ 2 A2 v 2 ,
1 mole of water = 18 grams.
p v2
+ +gy = constant on a streamline.
ρ 2
9
tC + 32, T = tC + 273.15, ∆L = αL∆T, ∆V = βV ∆T.
5
J
R
J
R = 8.314
, NA = 6.022×1023 particles in a mole, k =
= 1.381×10−23 .
mol · K
NA
K
tF =
P V = nRT, P V = N kT, N = nNA ,
r
3
1
3kT
mhv 2 i = kT, vrms =
, N hgav i = g1 + g2 + g3 + ... + gN .
2
2
m
Thermal radiation power : area A, P = σT 4 A, strongest wavelength, λmax = 2.9×10−3
Thermodynamics: For solids and liquids of mas m and specific heat capacity c, Qin =
cm∆T if there is no phase transition. During any phase transition (sublimation, melting,
evaporation) Qin = mL, where L is the latent heat of the transition.
Z
First Law : ∆U = Qin − Wovt ,
Vf
Wovt =
P dV.
Vi
For n moles of an ideal gas dU = nCV dT, QVin = nCV dT, QPin = nCP dT,
where c0 is the molar heat capacity. CP = CV + R. For a monatomic ideal gas CV = 3R/2
and for a diatomic gas CV = 5R/2.
In a heat engine, η =
Wnet
,
Qin
ηCarnot = 1 −
TC
,
TH
because in Carnot
QH
QC
=
.
TH
TC
Tf
Vf
Change of entropy in an ideal gas process, ∆S = nCV ln
+ nR ln
.
Ti
Vi
206 Sample Final
Spring 2010
m·K
.
T
Dr. Huerta
PHY 206 Sample Final
Spring 2010
Waves: Sinusoidal traveling wave p(x, t) = A sin(kx−ωt+φ), k = 2π/λ, ω = 2πf, v = λf.
Doppler effect:
Moving source : fobserved = fsource
1
1±
vsource
vwave
s
In a string v =
Intensity : I(r) =
, Moving observer : fobserved = fsource (1±
T
,
µ
vobserver
).
vwave
1 2 2
µA ω v.
2
I
β = 10 log
.
I0
hP i =
Psource
,
4πr2
Optics: In a medium with index of refraction n, the speed of light is
v=
Reflection : θ1 = θ2 ,
Mirrors :
c
,
n
where c = 3 × 108 m/s.
Ir
(n2 − n1 )2
. Refraction : n1 sin θ1 = n2 sin θ2 .
=
I0
(n2 + n1 )2
1 1
1
+ = ,
s
i
f
Single refracting surface :
R
. For plane mirror R = ∞.
2
n1
n2
n2
n2
+
=
,
f=
R.
s
i
f
n2 − n1
f=
For a thin lens, with the same external medium on both the left and right sides, and with
n the index of the lens material relative to the medium ovtside, n = nmaterial /nmedium ,
1 1
1
1
1
1
+ = ,
= (n − 1)
−
.
s
i
f
f
R1
R2
Two thin lenses of focal lengths f1 and f2 placed one very near behind the other have a
combined focal length f where
1
1
1
=
+ .
f
f1
f2
The near point dmin is the smallest distance from the eye where a person can see sharply.
The normal dmin = 25 cm. The far point dmax is the largest distance from the eye where
a person can see sharply. The normal dmax = ∞. For a simple magnifier of focal length
fm , used by a person with near point dmin , the angular magnification with a relaxed eye
(virtual image at ∞) is
Mθ =
dmin
fobjective
. For a telescope Mθ =
.
f
feyepiece
Interference: Superposition of two waves: Enet = E1 + E2 . With two coherent sources
that are oscillating ovt of phase by φ = φ2 − φ1 , we take E1 = E0 sin(kr1 − ωt + φ1 ), and
206 Sample Final
Spring 2010
Dr. Huerta
PHY 206 Sample Final
Spring 2010
E2 = E0 sin(kr2 − ωt + φ2 ) . Intensity is I = KhE 2 i, where hE 2 i is the time average.
2
I1 = I2 = cE02 /2 ≡ Io , and Inet ≡ Enet
/2, we get at the arrival point
Inet
φ
2π
= 4I0 cos
, where φ =
∆L+φ, is the phase difference of the waves at that point.
2
λ
2
max :
max : ∆L +
φ
= ±0, π, 2π, 3π, ...,
2
λ
φ = ±nλ, n = 0, 1, 2, 3...,
2π
min :
φ
π 3π 5π
= ±, ,
,
..., or
2
2 2 2
min := ∆L +
λ
1
δ = ±(n + )λ, n = 0, 1, 2, 3...
2π
2
In thin films ray 1 may be reflected with a phase shift of δ1 = 0, or δ1 = π. Ray 2 may also
may be reflected with a phase shift of δ2 = 0, or δ2 = π. Depending on this δ = δ2 − δ1
may be 0, or π.
Double slit interference: For slit separation d, and with m=0,1,2,...
d sin θ = mλ for maximum,
1
and d sin θ = (m + )λ for minimum.
2
If the distance from the slits to the observing screen is D d, the angles are small, and
y is measured from the central maximum (m=0),
max
the maxima are found at ym
= mλ
D
,
d
1
D
min
and the minima are found at ym
= (m+ )λ
.
2
d
Then the distance between successive maxima, or successive minima, is
∆y = λ
D
d
Diffraction When light falls with rays perpendicular to a single slit of width a, the
diffraction pattern is
I = Imax
sin α
α
2
, where α =
πa sin θ
. The mth minimum is at a sin θ = mλ, m = 1, 2, 3, ...
λ
When light falls with rays perpendicular to a circular aperture of diameter a, the first
minimum of the diffraction pattern is at
θmin = 1.22
λ
.
D
Resolution of Optical Instruments: Two point sources (such as two stars, or the
headlights on a car) at a distance s form the aperture can be resolved if their separation
w obeys w > sθmin .
206 Sample Final
Spring 2010
Dr. Huerta
PHY 206 Sample Final
Spring 2010
Diffraction Grating A diffraction grating with slit separation d and a total of N slits
produces a diffraction pattern
I = Imax
sin(N β)
sin β
2
where β =
πd sin θ
. The mth order maximum is at d sin θ = mλ, m = m1, 2, 3...
λ
Special Relativity For two inertial observers O and O0 , with O0 moving with velocity v
along the x axis relative to observer O, the Galilean transformations are
x0 = x − vt,
or also
x = x0 + vt0 ,
The Lorentz transformations are
vx
0
0
x = γ(x − vt), t = γ t − 2 ,
c
or also
0
0
x = γ(x +vt ),
y 0 = y,
0
0 vx
t=γ t+ 2 ,
c
t0 = t,
y 0 = y,
and z 0 = z,
y = y0 ,
and z 0 = z.
where
and z = z 0 ,
γ=q
where
1
1−
,
v2
c2
γ=q
1
1−
.
v2
c2
For an object with velocity u relative to O and u0 relative to O0 ,
u0 =
u−v
,
1 − uv
c2
or also
u=
u0 + v
.
0
1 + uc2v
Time dilation: For two events that happen at the same place as far as O0 is concerned, so
x01 = x02 (O0 is the proper observer), but as far as O is concerned O0 travels from event 1 to
event 2 with speed v, so x1 6= x2 (O is the improper observer), but rather, x2 = x1 + v∆t,
the time intervals between the events are related by
r
∆t0 = ∆t 1 −
v2
,
c2
so O0 (proper) measures a smaller time than O (improper).
Length contraction: Say that O says that the length of the trip was ∆x = LO = v∆t.
(The length LO is at rest in the O reference frame). As far as O0 is concerned
q the trip
took a short time ∆t0 because it was a short trip of length v∆t0 = LO0 = LO
measures a shorter length because the length in question
206 Sample Final
1−
v2
c2 .
O0
Spring 2010
Dr. Huerta
PHY 206 Sample Final
Spring 2010
[1.] This problem has five multiple choice questions. Circle the best answers.
[1A.] A red star is seen to have its brightest thermal radiation at a wavelength of 660 nm. The
red glowing surface of the star is at a temperature of
[A] 5800 K [B] 21,783 K [X] 4394 K [D] 1.05 × 106 K [E] 1.05 × 10−3 K
[1B.] A monochromatic point source radiates 120 W uniformly in all directions. What is the
intensity at a distance of 2 m?
[X] 2.4 W/m2 [B] 4.8 W/m2 [C] 30 W/m2 [D] 9.6 W/m2 [E] 6 × 103 W/m2
[1C.] A heat engine receives 12,000 J of heat from the burning fuel and releases 9,000 J to the
environment in each cycle. The efficiency of the engine is
[A] 3 [B] 4 [X] 1/4 [D] 1/3 [E] 3/4
[1D.] One mole of a monoatomic ideal gas is compressed isothermally until its final volume is
1/2 of the initial volume. After the gas reaches equilibrium the change of entropy is
[A] zero [X] −R ln 2 [C] R ln 2 [D] 23 R ln 2 [E]- 32 R ln 2
[1E.] A nearsighted man has a far point located only 200 cm from his eyes. Determine the power
of contact lenses that will enable him to see distant objects clearly.
[A] -2 diopters [B] 2 diopters [X] -0.5 diopters [D] 0.5 diopters [E] -200 diopters
206 Sample Final
Spring 2010
Dr. Huerta
PHY 206 Sample Final
Spring 2010
[2.] This problem has five multiple choice questions. Circle the best answers.
[2A.] When light of wavelength 500 nm falls perpendicularly on a single slit, the first two dark
diffraction fringes (one on each side of the central bright fringe) are separated by a distance
of 10 cm when projected on a screen 10 m away. How wide is the slit? (for small angles
sin θ ≈ tan θ)
[A] 4 × 10−4 m [B] 2 × 10−4 m [X] 10−4 m [D] 0.25 × 10−4 m [E] 0.5 × 10−4 m
[2B.] Coherent light falls on a pair of slits separated by 0.5 mm. The interference pattern is
observed on a screen 10 m behind the slits. The separation on the screen between adjacent
bright interference fringes is 10 mm. What is the wavelength of the light?
[A] 125 nm [B] 2000 nm [C] 250 nm [D] 1000 nm [X] 500 nm
[2C.] Two slits of width a and separation d are illuminated by coherent light. The central
diffraction envelope of the double-slit diffraction pattern contains 9 bright interference
fringes, and the first diffraction minima eliminate (are coincident with) bright interference
fringes. d and a obey
[X] ad = 5
[C] ad = 9
[D] ad = 5
[A] ad = 9
[E] the wavelength of the light is needed to answer
[2D.] The headlights on a car are separated by a distance w. A car is a distance s away from
you, and the pupil (opening) of your eye has a diameter a. If the light has wavelength λ
and you can resolve the two headlights, the distance s obeys
aw
aw
[A] s > 1.22λ
[B] s < 1.22λ
[C] s > 1.22λ
[X] s < 1.22λ
[E] s < 1.22λ
aw
aw
a
[2E.] Light of wavelength λ falls perpendicularly on a diffraction grating of width L with N slits.
Let Int(x) the integer part of x. For example if x = 4.6, Int(4.6) = 4. What is the largest
order maximum that is produced? (If sin θ ≥ 1 that order cannot be produced.)
Nλ
L
Nλ
[C] mlargest = N λ [X] mlargest = Int NLλ
[A] mlargest = L [B] mlargest = Int L
206 Sample Final
Spring 2010
Dr. Huerta
PHY 206 Sample Final
Spring 2010
[3.] This problem has five multiple choice questions. Circle the best answers.
[3A.] Two stars are at rest with respect to each other separated by a distance of 12 light-years
as measured by observers at rest with respect to the stars. Observers in the stars observe
an astronaut fly from one star to the other with a speed of v = 0.6c. How long did the
trip last as measured by the observers at rest with respect to the stars?
[A] 12 years. [B] 12.8 years. [X] 20 years [D] 16 years [E] 25 years.
[3B.] In the problem above, how long did the trip last as measured by the astronaut?
[A] 12 years. [B] 12.8 years. [C] 20 years [X] 16 years [E] 25 years.
[3C.] As measured by the astronaut, the distance between the stars in the problem above is
[A] 7.2 light-years
[X] 9.6 light-years
[C] 12 light-years
[D] 15 light-years
[E]
20 light-years
[3D.] An enemy spaceship is approaching us with a speed of 0.8c and fires a photon torpedo
toward us whose speed relative to the spaceship is c. What is the speed of the photon
torpedo relative to us?
[A] 0.8c [B] 1.8c [C] 0.2c [D] 1.3c [X] c
[3E.] A Particle is moving with in the x direction with a velocity of vi = 0.6c. If its velocity is
increased to vf = 0.8c, the ratio of its final momentum over its initial momentum is
q
p
p
p
p
p
9
[B] pfi = 43
[C] pfi = 34
[D] pfi = 16
[E] pfi = 16
[A] pfi = 43
9
206 Sample Final
Spring 2010
Dr. Huerta
PHY 206 Sample Final
Spring 2010
[4.] One mole of an ideal diatomic gas with molar specific heat Cv = 52 R is taken through the
cyclic process shown on the P − V diagram where P1 = 3P3 and V3 = 2V1 .
(a) Remembering the ideal gas law, the first law of Thermodynamics, the energy equation,
the expression for work in a quasistatic process
f
Z
P V = nRT,
∆U = ∆Q − ∆W,
∆U = nCv ∆T,
and
∆Wif =
P dV = Area,
i
and the entropy change in an arbitrary process,
∆Sif = nCV ln
Vf Tf + nR ln
,
Ti
Vi
fill in the table with the values of ∆U12 , ∆U23 , ∆U31 , ∆Unet , ∆W12 , ∆W23 , ∆W31 , ∆Wnet , ∆Q12 , ∆Q23 ,
∆Q31 , ∆Qnet , and ∆S12 , ∆S23 , ∆S31 , ∆Snet , expressing everything in terms of P1 , V1 , and
R.
(b) Calculate the efficiency of the cycle η.
P
2
1
1
P
3
3
V
!U
V3
1
! W
! Q
! S
1-2
2-3
3-1
total
Answers
[a] ∆Uif = nCV (Tf − Ti ) = 25 (Pf Vf − Pi Vi ), therefore
∆U12
5
5
5
5 −2
−10
= (P2 V2 − P1 V1 ) = P1 V1 , ∆U23 = (P3 V3 − P2 V2 ) =
P1 V 1 =
P1 V1 ,
2
2
2
2 3
3
∆U31 =
206 Sample Final
5
5
(P3 V3 − P1 V1 ) = P1 V1 , and ∆Unet = 0.
2
6
Spring 2010
Dr. Huerta
PHY 206 Sample Final
∆W12 = P1 (V3 − V1 ) = P1 V1 , ∆W23 = 0, ∆W31 =
and ∆Wnet =
Spring 2010
1
−2
(P1 − P3 )(V3 − V1 ) =
P1 V 1 ,
2
3
1
P1 V 1.
3
For the heat flows we use ∆Q = ∆U + ∆W .
∆Q12 = ∆U12 + ∆W12 =
7
−10
P1 V1 , ∆Q23 = ∆U23 + ∆W23 =
P1 V 1 ,
2
3
∆Q31 = ∆U31 + ∆W31 =
1
1
P1 V1 , and ∆Qnet = P1 V1 .
6
3
For ∆S we get
∆S12
∆S31
P2 V2
7R
V2
5
= nCV ln
ln 2,
+ nR ln
= R ln 2 + R ln 2 =
P1 V1
V1
2
2
P3 V3
1
−5R
V3
5
∆S23 = nCV ln
+ R ln(1) =
ln 3,
+ nR ln
= R ln
P2 V2
V2
2
3
2
P1 V 1
V1
3
1
5R
7R
5
= nCV ln
ln 3−
ln 2, and ∆Snet = 0.
+nR ln
= R ln( )+R ln( ) =
P3 V 3
V3
2
2
2
2
2
[c] The efficiency is
∆Wnet
η=
=
∆Qin
206 Sample Final
1
3 P1 V 1
7
1
2 P1 V1 + 6 P1 V1
=
1
1/3
=
.
22/6
11
Spring 2010
Dr. Huerta
PHY 206 Sample Final
Spring 2010
[5.] Observer O0 has a velocity v in the +x direction with respect to observer O.
[a] An object has a velocity u along the x axis relative to O and a different velocity u0 with
respect to O0 . Use the Lorentz transformations to derive the relativistic formula that gives
u0 in terms of u, v, and c.
[b] Events 1 and 2 have coordinates (x1 , t1 ) and (x2 , t2 ) as measured by O, and coordinates
(x01 , t01 ) and (x02 , t02 ) as measured by O0 . The two events are simultaneous as measured by
O, that is t1 = t2 . Find the time interval between the events as measured by O0 .
Answers:
[a] By definition
∆x
∆x0
, and u =
.
u0 =
0
∆t
∆t
Therefore
u0 =
∆x
γ(∆x − v∆t)
u−v
∆t − v
=
=
,
v
∆x
γ(∆t − v∆x/c2
1 − uv
1 − c2 ∆t
c2
u0 =
so
u−v
.
1 − uv
c2
[b] We use that ∆t = 0 because t1 = t2 , and
vx
t =γ t− 2 ,
c
0
therefore
v∆x
∆t = γ ∆t − 2 ,
c
0
v∆x
or ∆t = −γ
.
c2
0
So, if ∆x = x2 − x1 > 0, then ∆t0 = t02 − t01 < 0, and O0 measures that event 2 happened
before event 1. If instead ∆x < 0, then ∆t0 = t02 − t01 > 0, and O0 measures that event 2
happened later than event 1.
206 Sample Final
Spring 2010
Dr. Huerta
PHY 206 Sample Final
Spring 2010
[6.] A standing wave is established in a 100 cm long string fixed at both ends. The string
vibrates in a normal mode with 5 segments, with a node at each end and 4 intermediate
nodes, at a frequency of 50 Hz.
(a) What is the speed of the waves in the string?
(b) What is the fundamental frequency?
(c) Sketch the mode of vibration above and give its mathematical form for the displacement
y(x, t) if at t = 0 the string has y(x, 0) = 0.
Answers:
L = 100 cm
(a) Each segment is λ/2,
L=5·
λ
,
2
so
λ=
2
2
L = 100 cm = 40 cm,
5
5
and f = 50 Hz.
Bvt
v = λf
therefore
v = 0.4 m · 50 Hz = 20 m/s.
(b) The fundamental mode has L = λ/2, or λ = 2L = 2 m, and the same speed as above, so
f = v/λ = 10 Hz. The mode shown in the figure is the fifth harmonic.
(c) y(x, t) = y0 sin kx sin ωt because it must be zero at x = 0 and t = 0. Bvt k = 2π/λ, and
λ = 0.4 m, so k = 5π m−1 . Also, ω = kv = 5π m−1 · 20 m/s = 100π s−1 . Therefore
y(x, t) = y0 sin(5π m−1 x) sin(100π s−1 t).
206 Sample Final
Spring 2010
Dr. Huerta
PHY 206 Sample Final
Spring 2010
[7.] A 10 cm tall object is located 60 cm to the left of a lens whose focal length is 20 cm.
(a) Find the position, size, and type of the image. Accompany your calculations with an
approximate drawing of the rays.
(b) A second lens of 15 cm focal length is placed 50 cm to the right of the first lens. Find
the position, size, and type of the final image. Again, accompany your calculations with
an approximate drawing of the rays.
(c) Say the second lens is placed 10 cm, not 50 cm, to the right of the first lens. Find the
position, size, and type of the final image. Again, accompany your calculations with an
approximate drawing of the rays.
Answers:
(a) s1 = 60 cm, f1 = 20 cm, y1 = 10 cm,
1 1
1
1
2
s0
30
1 1
−
=
, so s01 = 30 cm, and y10 = − 1 y1 = − 10cm = −5cm.
+ 0 = , 0 =
s1 s1
f1 s1
20 cm 60 cm
60 cm
s1
60
The image I1 produced by the first lens is real and inverted, 5 cm tall, and 30 cm to the
right of the first lens. It serves as an object for the second lens.
f = 15 cm
2
f1 = 20 cm
O
s = 60 cm
1
I
y = 10 cm
1
F
1
F1
F
20 cm
I1
y’ = −5 cm = y
2
1
s’ = 30 cm
1
2
y’ = 15cm
2
2
s =20cm
2
F
2
s’ = 60 cm
2
(b) s2 = 50 cm − 30 cm = 20 cm, f2 = 15 cm, y2 = −5 cm.
1
1
1
s02
60
1
0
0
=
−
=
,
so
s
=
60
cm,
and
y
=
−
y2 = − (−5) = 15 cm.
2
2
0
s2
15 cm 20 cm
60 cm
s2
20
The image I2 produced by the second lens is real and upright, 15 cm tall and 60 cm to the
right of the second lens.
(c) Now I1 is 20 cm beyond the second lens. Light rays are converging toward it are interrupted
by the second lens. I1 serves as a virtual object for the second lens, with a negative object
distance s2 = −20 cm, f2 = 15 cm, y2 = −5 cm,
206 Sample Final
Spring 2010
Dr. Huerta
PHY 206 Sample Final
f = 20 cm
1
O
Spring 2010
f = 15 cm
2
s = 60 cm
1
y = 10 cm
1
F F
1 2
F2
F1
20 cm
I
1
2
I1
y’ = −5 cm = y
2
1
and y’ = 2.1 cm
2
s’ = 8.6 cm
2
s = −20 cm
2
s’ = 30 cm
1
10 cm
1
1
7
60
s02
60/7
15
1
0
0
−
=
,
so
s
=
cm,
and
y
=
−
y2 = −
(−5) =
cm = 2.1 cm.
=
2
2
0
s2
15 cm (−20) cm
60 cm
7
s2
(−20)
7
Now the image I2 produced by the second lens is real and inverted, 2.1 cm tall and 8.6
cm to the right of the second lens. To find the final image I2 graphically we use a first
ray through the vertex of the second lens that is aimed at I1 . I also drew how this ray
would have started at some peculiar angle from the object. This first ray continues straight
through the lens. Then we use a second ray parallel to the axis aimed at I1 . When this
second ray passes through the second lens it is bent through the focus. Where the first
and second rays intersect, we find I2 .
206 Sample Final
Spring 2010