Physics 112
Homework 10 (solutions)
(2004 Fall)
Solutions to Homework Questions 10
Chapt24, Problem-1:
A laser beam (! = 632.8 nm) is incident on two slits 0.200 mm apart. How far apart
are the bright interference fringes on a screen 5.00 m away from the double slits?
Solution:
You are asked about bright fringes, so we need constructive interference, so the path lengths taken by the
light going the two slits has to differ by an integer number of wavelengths. So considering two adjacent
fringes, namely the “mth-order” and the next one (m+1), we get
#L
#L
#L
(m + 1) " m =
d
d
d
"9
632.8 ! 10 m ( 5.00 m )
=
= 1.58 !10 "2 m = 1.58 cm
0.200 ! 10"3 m
!y bright = y m+1 " y m =
(
)
Chapt24, Problem-4: Light of wavelength 460 nm falls on two slits spaced 0.300 mm apart. What is the
required distance from the slit to a screen if the spacing between the first and second dark fringes is to be 4.00 mm?
Solution:
So this time you are dealing with dark fringes (destructive interference; odd 1/2 wavelength difference in
path length), and we are given the separation between the fringes, and asked to calculate the distance to
the screen.
!L "
1%
" L $ 3 1' " L
.
$ m + ' , the spacing between the first and second dark fringes is !y =
& # )=
#
&
2
d
d % 2 2(
d
(!y ) d 4.00 # 10$3 m 0.300 #10 -3 m 2 .61 m
Thus, the required distance to the screen is L =
=
=
"
460 # 10$9 m
From y dark =
(
)(
)
Chapt24, Problem-9: Waves from a radio station have a wavelength of 300 m. They travel by two paths to a
home receiver 20.0 km from the transmitter. One path is a direct path, and the second is by reflection from a mountain
directly behind the home receiver. What is the minimum distance from the mountain to the receiver that produces
destructive interference at the receiver? (Assume that no phase change occurs on reflection from the mountain.)
Solution:
The key here is that the “mountain” is directly behind the house, so the geometry is simple: the path
difference in the two waves received at the home is ! = 2 d , where d is the distance from the home to the
mountain. Neglecting any phase change upon reflection, the condition for destructive interference is
1%
"
! = $ m + ' ( with m = 0, 1, 2,… ,so the minimum distance to the mountain is half this distance, and
#
2&
for the first destructive interference fring (m=0) will occur when
1 $ ' ' 300 m
!
d min = # 0 + & = =
= 75.0 m
"
2% 2 4
4
(kinda small mountain I guess)
1
Physics 112
Homework 10 (solutions)
(2004 Fall)
Chapt24, Problem-14: A soap bubble (n = 1.33) is floating in air. If the thickness of the bubble
wall is 115 nm, what is the wavelength of the visible light that is most strongly reflected?
Solution:
Light reflecting from the first surface suffers phase reversal (since n2 > n1 ) Light reflecting from the
second surface does not, but passes twice through the thickness t of the film. The term “most-strongly
reflected” basically means we are looking for constructive interference (path difference is an integer
number of wavelengths). We are also making the assumption that the angle of incidence is approximately
zero-degrees. So we require
!n
!
= m! n , m = 1, 2,… , where !n = 0 is the wavelength in the film.
2
n film
4n film t 4 (1.33 )(115 nm ) 612 nm
1% (
"
Then 2 t = $ m ! ' 0 or !0 =
=
=
#
2 & n film
2m " 1
2m " 1
2m " 1
2t+
Of the possible wavelengths, only !0 = 612 nm , associated with m = 1 , is in the visible spectrum
Chapt24, Problem-21: Astronomers observe the chromosphere of the Sun with a filter that
passes the red hydrogen spectral line of wavelength 656.3 nm, called the H! line. The filter consists of a
transparent dielectric of thickness d held between two partially aluminized glass plates. The filter is held at a
constant temperature. (a) Find the minimum value of d that will produce maximum transmission of
perpendicular H! light, if the dielectric has index of refraction 1.378. (b) If the temperature of the filter
increases above the normal value, thereby increasing its thickness, what happens to the transmitted
wavelength? (c) The dielectric will also pass what near-visible wavelength? One of the glass plates is colored
red to absorb this light.
Solution:
(a) For maximum transmission, we want destructive interference in the light reflected from the front
and back surfaces of the film.
If the surrounding glass has refractive index greater than 1.378, light reflected from the front surface
suffers no phase reversal and light reflected from the back does undergo phase reversal. This effect by
itself would produce destructive interference, so we want the distance down and back to be one whole
wavelength in the film. Thus, we require that
2t = !n = !0 n film or
t=
!0
656.3 nm
=
= 238 nm
2 ( 1.378)
2n film
(b)
The filter will expand. As t increases in 2n film t = !0 , so does ! increase .
(c)
Destructive interference for reflected light happens also for !0 in 2 t = 2 !0 n film , or
!0 = n filmt = (1.378 )(238 nm ) = 328 nm (near ultraviolet)
2
Physics 112
Homework 10 (solutions)
(2004 Fall)
Chapt24, Problem-29: Helium-neon laser light (! = 632.8 nm) is sent through a 0.300-mm-wide
single slit. What is the width of the central maximum on a screen 1.00 m from the slit?
Solution:
The distance on the screen from the center to either edge of the central maximum is
$ #'
y = L tan ! " L sin ! = L & )
% a(
$ 632.8 * 10+9 m '
+3
= (1.00 m ) &
) = 2.11 * 10 m = 2.11 mm
+3
% 0.300 * 10 m (
The full width of the central maximum on the screen is then 2y = 4.22 mm
Chapt24, Problem-37: The hydrogen spectrum has a red line at 656 nm and a violet line at 434
nm. What is the angular separation between these two spectral lines obtained with a diffraction grating that
has 4 500 lines/cm?
Solution:
The grating spacing is d =
spectral lines will be
or
1 cm
1m
=
. From d sin ! = m" , the angular separation between the given
4500 4.50 !105
% m$ red (
#1 % m $ violet (
!" = sin #1 '
* # sin '
*,
& d )
& d )
% m 656 $ 10#9 m 4.50 $105 (
% m 434 $10 #9 m 4.50 $105 (
#1 '
*
*
!" = sin
# sin
'
1m
*
'
1m
*
&
)
&
)
The results obtained are: for m = 1, !" = 5.91° ; for m = 2, !" = 13.2° ; and for m = 3, !" = 26.5° .
Complete orders for m ! 4 are not visible.
#1 '
(
)(
)
(
)(
)
Chapt24, Problem-38: A grating with 1 500 slits per centimeter is illuminated with light of
wavelength 500 nm. (a) What is the highest-order number that can be observed with this grating? (b) Repeat
for a grating of 15 000 slits per centimeter.
Solution:
1 cm
= 6.67 !10 "4 cm = 6.67 ! 10"6 m , the highest order of ! = 500 nm that can be
1500
#6
d sin 90° 6.67 "10 m ( 1)
observed will be
m max =
=
= 13.3 or 13 orders
!
500 " 10#9 m
1 cm
(b)
If d =
= 6.67 !10 "5 cm = 6.67 ! 10"7 m , then
15 000
(a)
If d =
(
(
)
)
#7
d sin 90° 6.67 "10 m ( 1)
m max =
=
= 1.33 or 1 order
!
500 " 10#9 m
3
Physics 112
Homework 10 (solutions)
(2004 Fall)
Chapt24, Problem-45: The angle of incidence of a light beam in air onto a reflecting surface is
continuously variable. The reflected ray is found to be completely polarized when the angle of incidence is
48.0°. (a) What is the index of refraction of the reflecting material? (b) If some of the incident light (at an
angle of 48°) passes into the material below the surface, what is the angle of refraction?
Solution:
(a) From Brewster’s law, the index of refraction is n2 = tan !p = tan ( 48.0° ) = 1.11
(b) From Snell’s law, n2 sin !2 = n1 sin !1 , we obtain when !1 = ! p
# n 1 sin !p &
"1 # ( 1.00 ) sin 48.0° &
!2 = sin "1 %
( = 42.0°
( = sin %
'
$
1.11
$ n2 '
Note that when !1 = ! p , !2 = 90.0° " ! p as it should.
Chapt24, Problem-53: Three polarizing plates whose
planes are parallel are centered on a common axis. The
directions of the transmission axes relative to the common
vertical direction are shown in Figure P24.53. A linearly
polarized beam of light with the plane of polarization parallel to
the vertical reference direction is incident from the left on the
first disk with intensity Ii = 10.0 units (arbitrary). Calculate the
transmitted intensity If when !1 = 20.0°, !2 = 40.0°, and !3 =
60.0°. (Hint: Make repeated use of Malus’s law.)
Solution:
From Malus’s law, the intensity of the light transmitted by the first polarizer is I 1 = I i cos 2 !1 . The plane
of polarization of this light is parallel to the axis of the first plate, and is incident on the second plate.
Malus’s law gives the intensity transmitted by the second plate as
I 2 = I 1 cos 2 ( !2 " !1 ) = I i cos2 !1 cos 2 ( !2 " !1 ) . This light is polarized parallel to the axis of the second plate
and is incident upon the third plate. A final application of Malus’s law gives the transmitted intensity as
I f = I 2 cos2 (!3 " !2 ) = I i cos2 !1 cos 2 ( !2 " !1 ) cos 2 ( !3 " !2 )
With !1 = 20.0°, !2 = 40.0°, and !3 = 60.0° , this result yields
I f = (10.0 units ) cos 2 ( 20.0° ) cos 2 ( 20.0° ) cos 2 ( 20.0° ) = 6.89 units
Chapt24, Problem-55: Light with a wavelength in vacuum of 546.1 nm falls perpendicularly on a
biological specimen that is 1.000 µm thick. The light splits into two beams polarized at right angles, for
which the indices of refraction are 1.320 and 1.333. (a) Calculate the wavelength of each component of the
light while it is traversing the specimen. (b) Calculate the phase difference between the two beams when
they emerge from the specimen.
Solution:
(a)
If light has wavelength !0 in vacuum, its wavelength in a medium of refractive index n is ! = !0 n .
Thus, the wavelengths of the two components in the specimen are
!1 =
(b)
!0 546.1 nm
=
= 413.7 nm
1.320
n1
and
t 1.000 "10 -6 m
=
= 2.417
!1 413.7 "10 -9 m
and
!2 =
!0 546.1 nm
=
= 409.7 nm
1.333
n2
The number of cycles of vibration each component completes while passing through the specimen
t 1.000 "10 -6 m
=
= 2.441
!2 409.7 "10 -9 m
Thus, when they emerge, the two components are out of phase by N 2 ! N 1 = 0.024 cycles . Since each cycle
are
N1 =
N2 =
represents a phase angle of 360°, they emerge with a phase difference of
!" = ( 0.024 cycles ) ( 360° cycle ) = 8.6°
4
Physics 112
Homework 10 (solutions)
(2004 Fall)
Chapt24, Conceptual-1: Your automobile has two headlights. What sort of interference pattern do you
expect to see from them? Why?
Solution:
You will not see an interference pattern from the automobile headlights for two reasons. The first is that
the headlights are not coherent sources and are therefore incapable of producing sustained interference.
Also, the headlights are so far apart in comparison to the wavelengths emitted that, even if they were
made into coherent sources, the interference maxima and minima would be too closely spaced to be
observable.
Chapt24, Conceptual-2: Holding your hand at arm’s lengtgh, you can readily block sunlight from
youur eyes, Why can you not block sound from your ears in this way?
Solution:
The wavelength of light is extremely small in comparison to the dimensions of your hand, so the diffraction
of light around obstacles the size of your hand is totally negligible. However, sound waves have wavelengths
that are comparable to the dimensions of the hand or even larger. Therefore, significant diffraction of
sound waves occurs around hand sized obstacles.
Chapt24, Conceptual-6: If white light is used in Young’s double-slit experiment rather than
monochromatic light, how does the interference patter change?
Solution:
Every color produces its own interference pattern and we see them superposed. The central maximum is
white. The first maximum is a full spectrum with violet on the inside and red on the outside. The second
maximum is also a full spectrum, with red in it overlapping with violet in the third maximum. At larger
angles, the light soon starts mixing to white again.
Chapt24, Conceptual-10: Often fingerprints left on a piece of glass such as a window pane show colored
spectra like that from a diffraction grating. Why?
Solution:
The skin on the tip of a finger has a series of closely spaced ridges and swirls on it. When the finger
touches a smooth surface, the oils from the skin will be deposited on the surface in the pattern of the
closely spaced ridges. The clear spaces between the lines of deposited oil can serve as the slits in a crude
diffraction grating and produce a colored spectrum of the light passing through or reflecting from the
glass surface.
Chapt24, Conceptual-19: When you receive a chest X-ray at a hospital, the X-rays pass through a series of
parallel ribs in your chest. Do the ribs act as a diffraction grating for X-rays?
Solution:
Strictly speaking, the ribs doo act as a diffraction grating, butt the separation distance oof the ribs is so
much larger than the wavelength of the X-rays that there are no observable effects.
5