1. No category

Dr. Huerta
PHY 206 Sample Test 3
Spring 2009
Spring 2009, Section FJ, MWF 1:25 – 2:15 p. m.
Name:
Signature:
1
Idnumber:
Z
p = mv,
2
f
Impulse =
Fdt = ∆p = Fav ∆t.
i
m
Use g = 10 2 ,
s
dp
= −ρg,
dy
ρwater
kg
= 1 × 103 3 ,
m
3
Patm = 105 Pa,
1 mole of water = 18 grams.
4
2
p(y) = p0 −ρgy,
ρ 1 A1 v 1 = ρ 2 A2 v 2 ,
p v
+ +gy = constant on a streamline.
ρ 2
5
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
1
3
3kT
mhv 2 i = kT, vrms =
, N hgav i = g1 + g2 + g3 + ... + gN .
2
2
m
Thermal radiation:
total power, area A,
P = σT 4 A,
wavelength with most power, λmax = 2.898×10−3
Z
∆U = Qin − Wout ,
m·K
.
T
Vf
Wout =
P dV.
Vi
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.
For n moles of an ideal gas dU = nc0V dT, QVin = nc0V dT, QPin = nc0P dT,
where c0 is the molar heat capacity. c0P = c0V + R. For a monatomic ideal gas c0V = 3R/2
and for a diatomic gas c0V = 5R/2.
In a heat engine, η =
Wnet
,
Qin
ηCarnot = 1 −
TC
,
TH
because in Carnot
Change of entropy in an ideal gas process, ∆S =
Sinusoidal traveling wave p(x, t) = A sin(kx − ωt + φ),
206 Sample Test 2
nc0V
Tf
ln
Ti
k = 2π/λ,
QH
QC
=
.
TH
TC
Vf
+ nR ln
.
Vi
ω = 2πf,
v = λf.
Spring 2009
Dr. Huerta
PHY 206 Sample Test 3
Spring 2009
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
,
r2
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.
(n2 − n1 )2
Ir
=
. 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
n2
n2
n2
n1
+
=
,
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 outside, n = nmaterial /nmedium ,
1
1 1
+ = ,
s
i
f
1
1
1
= (n − 1)
−
.
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 =
206 Sample Test 2
dmin
fobjective
. For a telescope Mθ =
.
f
feyepiece
Spring 2009
Dr. Huerta
PHY 206 Sample Test 3
Spring 2009
Interference: Superposition of two waves: Enet = E1 + E2 . With two coherent sources
that are oscillating out of phase by δ = δ2 − δ1 , we take E1 = E0 sin(kr1 − ωt + δ1 ), and
E2 = E0 sin(kr2 − ωt + δ2 ) . 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
2π
φ
, where φ =
∆L+δ, is the phase difference of the waves at that point.
= 4I0 cos
2
λ
2
φ
π 3π 5π
= ±, ,
,
..., or
2
2 2 2
λ
λ
1
max : ∆L + δ = ±nλ, n = 0, 1, 2, 3..., min := ∆L + δ = ±(n + )λ, n = 0, 1, 2, 3...
2π
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.
max :
206 Sample Test 2
φ
= ±0, π, 2π, 3π, ...,
2
min :
Spring 2009
Dr. Huerta
PHY 206 Sample Test 3
Spring 2009
[1.] This problem has five multiple choice questions. Circle the best answers.
[1A.] A thin lens is placed very near to the front of the eye of a person that has a near point of
1 meter. What focal length should the lens have for the person to be able to see an object
located 25 cm in front of the eye?
[X] 33 cm [B] 20 cm [C] -33 cm [D] -20 cm
1
2
glass
d
air
t
glass
x
D
[1B.] Light falls almost vertically as shown. When looked at from above, dark and bright
interference fringes appear in the air film between the two glass slabs. Let n = 0, 1, 2...
Bright fringes appear where
[A] t = nλ
[B] 2t = nλ
[C] t = (n + 12 )λ
[X] 2t = (n + 21 )λ
[1C.] If in the double slit experiment, the separation between the slits is doubled, the distance
between adjacent interference maxima
[A] is also doubled [B] is quadrupled [X] is reduced to half the original
[D] is reduced to one fourth of the original [E] remains the same
[1D.] Two coherent light sources, each with intensity of 1.0 × 103 W/m2 , interfere at a point at
which the phase difference φ = 60◦ . The intensity at this point is
[A] 1.0 × 103 W/m2 [B] 2.0 × 103 W/m2 [X] 3.0 × 103 W/m2
[D] 4.0 × 103 W/m2 [E] zero
[1E.]
[1D.] Two thin lenses of focal lengths f1 and f2 are placed one very near behind the other. The
focal length of the combination is
[A] f = f1 + f2 [B] f = f1 − f2 [C] f =
206 Sample Test 2
f1
f2 (f1
+ f2 ) [X]
1
f
=
1
f1
+
1
f2
[E]
1
f
=
1
f1
−
1
f2
Spring 2009
Dr. Huerta
PHY 206 Sample Test 3
Spring 2009
[2.] A light ray goes from point A to point B after reflecting from a mirror at point M. Show
that the path AMB will be a path of minimum time if the ray obeys the law of reflection.
A
B
s
M
mirror
x
d
Answers:
p
√
The distance from A to M is s2 + x2 , and from M to B it is s2 + (d − x)2 . The time
t(x) to travel the path AMB with speed v is
√
t(x) =
s2 + x2
+
v
p
s2 + (d − x)2
v
The time will be a minimum where
dt(x)
x
(−1)(d − x)
=0= √
+ p
.
dx
v s2 + x2
v s2 + (d − x)2
Therefore
√
x
(d − x)
,
=p
s2 + x2
s2 + (d − x)2
or
sin θ1 = sin θ2 , or θ1 = θ2 ,
where θ1 and θ2 are the angles the rays make with the normal to the mirror.
206 Sample Test 2
Spring 2009
Dr. Huerta
PHY 206 Sample Test 3
Spring 2009
[2.] A ray of red light in air falls on a prism with index of refraction n = 1.5 as shown. The
dashed lines represent normals. Find the angles α, β, γ, and δ if the incoming ray is parallel
to the bottom face of the prism.
74
0
!
"
#
53
0
$
53
0
Answers:
From the picture we see that α = 90◦ − 53◦ = 37◦ . From Snell’s law,
1 × sin 37◦ = 1.5 × sin β, therefore sin β = 0.4012, or β = 23.65◦ .
Then, from the triangle at the top, 74◦ + (90◦ − β) + (90◦ − γ) = 180◦ , or γ = 74◦ − β, or
γ = 74◦ − 23.65◦ = 50.35◦ , so
1.5 × sin γ = 1 × sin δ, or sin δ = 1.5 × 0.77 = 1.15 > 1,
which is impossible, δ does not exist, there is total internal reflection.
206 Sample Test 2
Spring 2009
Dr. Huerta
PHY 206 Sample Test 3
Spring 2009
[3.] An object located 50 cm in front of a lens produces a real image at a screen located behind
the lens. The lens is moved toward the object without moving the object nor the screen.
When the lens is 40 cm from the object a real image is again produced on the screen. Find
the distance from the object to the screen and the focal length of the lens.
Answer:
Say that the distance from the object to the screen is D. Then, in the first case the image
distance is i = D − 50 cm, so using
1
1
1
1
1 1
+ = , we get
+
= .
s
i
f
50 cm (D − 50 cm)
f
In the second case we still have the same D, but now s = 40 cm, and i = D − 40 cm, so
1
1
1
+
= .
40 cm (D − 40 cm)
f
Therefore we can solve for D from
1
1
1
1
+
=
+
,
50 cm (D − 50 cm)
40 cm (D − 40 cm)
which gives D = 90 cm. After that we find that f = 200/9 cm.
206 Sample Test 2
Spring 2009
Dr. Huerta
PHY 206 Sample Test 3
Spring 2009
[4.] A Young’s double slit experiment is performed with monochromatic light. The separation
between the slits is 0.5 mm, and the interference pattern on a screen 3 m away shows the
distance from one maximum to the next adjacent maximum to be 3 mm. What is the
wavelenght of the light? What would happen to the distance between the maxima if the
slit separation were doubled?
Answers:
Since we have that D d, the position of the nth maximum occurs at an angle θn , where
θ is very small, with d sin θn = nλ, where n = 0, ±1, ±2, .... From the figure we see that
yn
D
nλ
= sin θn ≈ tan θn =
, or yn = nλ .
d
D
d
The distance from one maximum to the next one is
∆y = yn+1 −yn = λ
D
d
0.5 mm
. Therefore λ = ∆y = 3 mm
= 5×10−4 mm = 5×10−7 .
d
D
3, 000 mm
So the wavelength is 500 nm. The distance between maxima gets smaller by a factor of
two when d is doubled because ∆y is inversely proportional to d.
plane
wavefronts
y
d
θ
D
double slit screen
observation
screen
206 Sample Test 2
Spring 2009
Dr. Huerta
PHY 206 Sample Test 3
Spring 2009
[5.] A concave mirror has a radius of curvature of 60 cm. Calculate the image position and
magnification of an object placed at a distance of 60 cm in front of the mirror.
Answers:
here s = 60 cm, and f = R/2 = 30cm. Then, using
1 1
1
1
1
1
1
+ = , we get
=
−
=
, so i = 60 cm.
s
i
f
i
30 cm 60 cm
60 cm
The image is real. The lateral magnification is
i
M = − = −1.
s
The image is inverted and the same size as the object.
206 Sample Test 2
Spring 2009