Physics 2130
Final Examination
Spring 2006
1- A transverse sinusoidal wave is generated at one end of a long, horizontal string by a
bar that moves up and down through a distance of 1.6 cm. The motion is continuous and
is repeated regularly 150 times per second. The string has linear density 220 g/m and is
kept under a tension of 140 N. Find the maximum value of (a) the transverse speed u and
(b) the transverse acceleration and (c) the wavelength of the wave.
y  y m cos(kx  t )
y m  0.8cm;
  300 rad / s, v  140/0.22 m / s  25.2m / s
u m  y m  7.5m / s; a m   2 y m  7106m / s 2 ;   v/f  25.2/(300 )m  2.7cm
2- A jet plane passes over you at a height of 5070 m and a speed of Mach 1.39. (a) Find
the Mach cone angle. (b) How long after the jet passes directly overhead does the shock
wave reach you? Use 331 m/s for the speed of sound.
sin(  )  1 / 1.39;
  46 o
d  5070 / tan( 46)  4895m
t  d / v s  10.6s
3- A tuning fork of unknown frequency makes 3.00 beats per second with a standard fork
of frequency 384 Hz. The beat frequency decreases when a small piece of wax is put on a
prong of the first fork. What is the frequency of this fork?
Since the beat frequency equals the difference between the frequencies of the two tuning
forks, the frequency of the first fork is either 381 Hz or 387 Hz. When mass is added to
this fork its frequency decreases (recall, for example, that the frequency of a mass-spring
oscillator is proportional to 1/ m ). Since the beat frequency also decreases the
frequency of the first fork must be greater than the frequency of the second. It must be
387 Hz.
Physics 2130
Final Examination
Spring 2006
4- The maximum electric field 10 m from an isotropic point source of light is 2.0 V/m.
What are (a) the maximum value of the magnetic field and (b) the average intensity of the
light there? (c) What is the power of the source?
(a) The magnetic field amplitude of the wave is
Bm 
Em
2.0 V / m

 6.7  109 T.
8
c
2.998  10 m / s
(b) The intensity is
b
g
hc
2
2.0 V / m
E2
I m 
 5.3  103 W / m2 .
7
8
2  0c 2 4   10 T  m / A 2.998  10 m / s
c
h
(c) The power of the source is
P  4r 2 I avg  4 10m  5.3 103 W/m2   6.7 W.
2
5- Unpolarized light is sent into a system of two polarizing sheets. The angles θ1 and θ2
of the polarizing directions of the sheets are measured counterclockwise from the positive
direction of the y axis (they are not drawn to scale in the figure). Angle θ1 is fixed but
angle θ2 can be varied. Figure below gives the intensity of the light emerging from sheet
2 as a function of θ2. (The scale of the intensity axis is not indicated.) What percentage of
the light's initial intensity is transmitted by the two-sheet system when θ2 = 90°? 36
Physics 2130
Final Examination
Spring 2006
We examine the point where the graph reaches zero:  2 = 160º. Since the polarizers
must be “crossed” for the intensity to vanish, then 1 = 160º – 90º = 70º. Now we
consider the case  2 = 90º (which is hard to judge from the graph). Since 1 is still equal
to 70º, then the angle between the polarizers is now  =20º. Accounting for the
“automatic” reduction (by a factor of one-half) whenever unpolarized light passes
1
through any polarizing sheet, then our result is 2 cos2() = 0.442  44%.
6- A 2.00-m-long vertical pole extends from the bottom of a swimming pool to a point
50.0 cm above the water. Sunlight is incident at angle θ = 55.0°. What is the length of the
shadow of the pole on the level bottom of the pool? nwater=**** 49
Consider a ray that grazes the top of the pole, as shown in the diagram that follows.
. m. The length of the shadow is x + L.
Here 1 = 90° –  = 35°, 1  0.50 m, and  2  150
x is given by
x  1 tan  1  (0.50 m) tan 35  0.35 m.
According to the law of refraction, n2 sin 2 = n1 sin 1. We take n1 = 1 and n2 = 1.33
(from Table 33-1). Then,
 2  sin 1
F
sin  I
Fsin 35.0IJ 2555
 sin G
G
J
K . .
.
Hn K H133
1
2
1
Physics 2130
Final Examination
Spring 2006
L is given by
L   2 tan  2  (150
. m) tan 25.55  0.72 m.
The length of the shadow is 0.35 m + 0.72 m = 1.07 m.
7- A sand grain is 3.00 cm from thin lens 1, on the central axis through the two
symmetric lenses. The distance between focal point and lens is 4.00 cm for both lenses;
the lenses are separated by 8.00 cm. (a) What is the distance between lens 2 and the
image it produces of the sand grain? Is that image (b) to the left or right of lens 2, (c) real
or virtual, and (d) inverted relative to the sand grain or not inverted? 133
(a) Our first step is to form the image from the first lens. With p1 = 3.00 cm and f1 =
+4.00 cm, Eq. 34-9 leads to
1 1 1
   i1  12.0cm.
p1 i1 f1
The corresponding magnification is m1 = –i1/p1 = 4. This image serves the role of
“object” for the second lens, with p2 = 8.00 + 12.0 = 20.0 cm, and f2 = –4.00 cm. Now,
Eq. 34-9 leads to
1 1 1
   i2  3.33 cm .
p2 i2 f 2
(b) The fact that i2 is negative means that the final image is virtual (and therefore to the
left of the second lens).
(c) The image is virtual.
(d) With m2 = –i2/p2 = 1/6, the net magnification is m = m1m2 = 2/3 > 0. The fact that m is
positive means that the orientation of the final image is the same as the (original) object.
Therefore, the image is not inverted.
Physics 2130
Final Examination
Spring 2006
8- A broad beam of light of wavelength 683 nm is sent directly downward through the
top plate of a pair of glass plates. The plates are 120 mm long, touch at the left end, and
are separated by 48.0 mm at the right end. The air between the plates acts as a thin film.
How many bright fringes will be seen by an observer looking down through the top
plate? 69
Consider the interference of waves reflected from the top and bottom surfaces of the air
film. The wave reflected from the upper surface does not change phase on reflection but
the wave reflected from the bottom surface changes phase by  rad. At a place where the
thickness of the air film is L, the condition for fully constructive interference
is 2 L  m  21  where  (= 683 nm) is the wavelength and m is an integer. This is
satisfied for m = 140:
b g
bm  g  b140.5gc683  10 mh 4.80  10
L
9
1
2

5
2
m = 0.048 mm.
At the thin end of the air film, there is a bright fringe. It is associated with m = 0. There
are, therefore, 140 bright fringes in all.
9- (a) In a double-slit experiment, what ratio of d to a causes diffraction to eliminate the
fourth bright side fringe? (b) What other bright fringes are also eliminated? 31
(a) In a manner similar to that discussed in Sample Problem 36-5,
d sin(  )  4
a sin(  )  1,
we find the ratio should be d/a = 4. Our reasoning is, briefly, as follows: we let the
location of the fourth bright fringe coincide with the first minimum of diffraction pattern,
and then set sin  = 4/d = /a (so d = 4a).
(b) Any bright fringe which happens to be at the same location with a diffraction
minimum will vanish. Thus, if we let
sin  = m1/d = m2/a = m1/4a,
or m1 = 4m2 where m2  1, 2, 3, . The fringes missing are the 4th, 8th, 12th, and so on.
Hence, every fourth fringe is missing.
Physics 2130
Final Examination
Spring 2006
10- Show that if a spherical refracting surface with an index of refraction n2 placed in a
medium with an index of refraction n1 the following expression holds:
Here p is the object distance, i is the image distance and r is the radius of the spherical
refracting surface. Assume rays make small angles with the central axis of the surface.
See page 945 (derivation of eq 34-8)
11- Standing waves are produced by the interference of two traveling sinusoidal waves,
each of frequency 100 Hz. The distance from the 2nd node to the 5th node is 60 cm. The
wavelength of each of the two original waves is:
(5  2) / 2  60
50 cm
40 cm
30 cm
20 cm
15 cm
12- A 40-cm long string, with one end clamped and the other free to move transversely, is
vibrating in its fundamental standing wave mode. If the wave speed is 320 cm/s the
frequency is:
  4(40)cm  160cm
f  v /   2 Hz
32 Hz
16 Hz
8 Hz
4 Hz
2 Hz
13- If the sound level is increased by 10 db the intensity increases by a factor of:
  log(
I
)
I0
 2  1  10
log(
I2
I
)  log( 1 )  10
I0
I0
10  log(
I2 / I0
)  log( I 2 / I 1 )  I 2 / I 1  1010
I1 / I 0
Physics 2130
Final Examination
Spring 2006
2
5
10
20
100
14- Two pipes are each open at one end and closed at the other. Pipe A has length L and
pipe B has length 2L. Which harmonic of pipe B matches in frequency the fundamental
of pipe A?
f  v /   2 Hz
f A  v /( 4 L)
f B  nv /(8L)
n  1,3,5...
The second
The third
There are none
The fourth
The fundamental
14- Light of uniform intensity shines perpendicularly on a totally absorbing surface, fully
illuminating the surface. If the area of the surface is decreased:
pr  I / c
Fr  p r A
1- the radiation pressure decreases and the radiation force decreases
2- the radiation pressure increases and the radiation force decreases
3- the radiation pressure increases and the radiation force increases
4- the radiation pressure stays the same and the radiation force increases
5- the radiation pressure stays the same and the radiation force decreases
Physics 2130
Final Examination
Spring 2006
15- The index of refraction of benzene is 1.80. The critical angle for total internal
reflection, at a benzene-air interface, is about:
sin(  )  1 / 1.8
1- 56°
2- 47°
3- 34°
4- 22°
5- 18°
16- An erect object is placed on the central axis of a thin lens, further from the lens than
the magnitude of its focal length. The magnification is +0.4. This means:
1- the image is virtual and erect, and the lens is a diverging lens
2- the image is real and inverted and the lens is a converging lens
3- the image is virtual and erect, and the lens is a converging lens
4- the image is virtual and inverted and the lens is a diverging lens
5- the image is real and erect and the lens is a converging lens
17- The bellows of an adjustable camera can be extended so that the largest film to lens
distance is one and one-half times the focal length. If the focal length is 12 cm, the
nearest object which can be sharply focused on the film must be what distance from the
lens?
1 1 1
 
 p  36cm
18 p 12
1- 12 cm
2- 24 cm
3- 36 cm
4- 48 cm
5- 72 cm
Physics 2130
Final Examination
Spring 2006
18- In a Young's double-slit experiment, a thin sheet of mica is placed over one of the
two slits. As a result, the center of the fringe pattern (on the screen) shifts by an amount
corresponding to 30 dark bands. The wavelength of the light in this experiment is 480 nm
and the index of the mica is 1.60. The mica thickness is:
L
N 2  N1

n2  n1
L
30
(480  10 6 mm)
1.6  1
1- 0.090 mm
2- 0.012 mm
3- 0.014 mm
4- 0.024 mm
5- 0.062 mm
19- Waves from two slits are in phase at the slits and travel to a distant screen to produce
the third side maximum of the interference pattern. The difference in the distance traveled
by the waves is:
L  d sin(  )  n
n  0,1,2,3...
For the third maximum L  2
1- half a wavelength
2- a wavelength
3- three halves of a wavelength
4- two wavelengths
5- five halves of a wavelength
20- Monochromatic light is normally incident on a diffraction grating that is 1 cm wide
and has 10,000 slits. The first order line is deviated at a 30° angle. What is the
wavelength, in nm, of the incident light?
d sin(  )  n
n  0,1,2,3...
0.01m
(0.5)    500nm
10000
1- 300
2- 400
3- 500
Physics 2130
Final Examination
Spring 2006
4- 600
5- 1000