Physics 214 Spring 99|Problem Set 8|Solutions
Handout April 1 1999
1. Reading Assignment and Computer Lab 2
3. Double Slits
A double slit is illuminated by light of wavelength 600 nm and produces
an interference pattern on a screen. A very thin slab of int glass (n =
1.65) is placed over only one of the slits. As a consequence, the central
maximum of the pattern moves to the position originally occupied by the
tenth order maximum. Find the thickness of the glass slab.
Reading Assignment:
Week Beginning March 15: Serway 37.3, 38.4
Week Beginning March 29: Serway 37.5, 37.6, 38.5, 38.1, 38.2, 38.3
Computer Lab 2, huygens, will be due to your section TA before Thursday, April 8. The following sections from the lab manual need to be
completed:
Section H1a : Thin Slits, Constructive Interference (Omit sections
H1b and H1c)
Section H2 : Single Slit Diraction
Section H3 : Double Slit Diraction
Before starting, please read the Introduction and Introduction to the
Simulation sections
SOLUTION: To nd the phase change caused by the insertion of the
glass slab, consider a segment of the the light's path of length t. The
phase change experienced by the light in traveling this distance is given
by 2t=. Now insert a glass slab of thickness t into this space; the phase
change of the light traversing the slab is now 2t=glass , in which the
wavelength of the light in the glass is glass = =n. Thus, the total phase
shift between the situation with no glass, and the one with the glass
in place, is
= 2t= , 2tn= = ,2t(n , 1)=:
With no glass in place the tenth order maximum (m=10) occurs when
d sin = m = 10, which corresponds to a phase shift of 2d sin =.
Since the phase shift due to the glass exactly cancels the phase shift at
m=10, we must have that 2(n , 1)t= = 2d sin = or t = d sin =(n , 1).
But we know that sin = 10=d so t = 10=(n , 1). Plugging in the
numbers we get t = 9:2m.
2. Intensity from a Double Slit
Yellow light from the mercury spectrum ( = 579 nm) illuminates a pair
of vertical slits separated by 0.15 mm. An interference pattern is produced
on a screen 2.5 m from the slits. Find the intensity of the light at a point
P , a distance 2.0 cm to the right of the central maximum, relative to the
intensity at the central maximum.
4. Double Slit Interference|Intensity
Consider a conventional double-slit interference experiment, but where
one of the slits is wider than the other. The intensity of the light reaching
the screen from the wide slit alone is 4I0, while that from the narrow slit
alone is I0 . Use phasors to compute the intensity of the light on the screen
when both slits are open. Express your result in terms of I0 and , the
phase dierence between waves coming from the two slits.
SOLUTION: At the point P , 2:0 cm from the central maximum, the angle
is sin = 2:0 cm=2:5 m so = 8 mrad. The path dierence from the two
slits to the point P is d sin = 1:2 m. The wavelength is 579 nm so
the path dierence corresponds to 2.07 wavelengths or a phase shift of 13
radians. Therefore, at the point P we are adding two waves that dier
in phase by 13 radians and the intensity of the sum of the two waves is:
Imax cos2 (=2) = (0:95)Imax where Imax is the intensity of the central
maximum.
SOLUTION: The intensity of light at a point is proportional to the square
of the amplitude of the electric eld at that point: I / E 2 . Thus if E0
is the amplitude at the screen for light coming from the narrow slit, then
the amplitude for light coming from the wide slit must be 2E0 , so that the
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intensity from this slit is 4I0 . The phasor that results from superimposing
the two waves is
y
S
2
d
S
1
where is the phase dierence between waves coming from the two
slits. From the law of cosines, the amplitude for the combined elds is
2 = E 2 (5 + 4 cos ), and so the intensity is
given by Etot
0
x
Itot = I0 (5 + 4 cos ):
They emit electromagnetic radiation in phase with each other and of
equal intensity. The wavelength is . In the analysis below, you may
neglect the fallo of intensity with distance from the point sources.
a) Find the positions of the maxima of the radiation signal seen as one
moves out from S1 along the positive x axis. Express your results in terms
of d and .
and S2 are two coherent point sources of radiation that are a distance d
apart along the y axis:
SOLUTION: The
a point x
p path dierence between the two sources to maximum
on the x axis is d2 + x2 , x. This must be a multiple of for
p2 2
constructive interference, and so maxima will occur when d + x =
x + m for any integer m. Squaring and rearranging we obtain
5. Interference
S1
for m an integer.
2 m2 2
xat max = d ,2m
b) Apply your result to nd the positions of all of the maxima along the
positive x axis in the case when d = 4 m and = 1 m.
SOLUTION: Substituting in the formula above, we nd that maxima
occur for x's of 7.5, 3, 1.17, and 0 meters.
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6. Antenna Array
radio signal disappears? What is the smallest angle (other than
zero) for which the radio signal is a maximum?
a) Two antennae are located as shown in the drawing below. They radiate
equal intensity radio waves with a wavelength which is smaller than the
distance between the antennae. Draw a phasor diagram labeled with
the dierence in phase (in terms of ) between the two phasors. What is
the smallest angle for which the radio signal (detected by a distant
receiver) disappears?
SOLUTION:
The signal disappears when the phasors add to zero; that is, when
= 23 , and sin = 32L . It's a maximum when the phasors all point in
the same direction again; that is, when = 2, and sin = 2L
SOLUTION:
The signal disappears when the phasors add to zero; that is, when
= , and sin = 2L
b) Now three antennae which radiate equal intensity radio waves with a
wavelength are located as shown in the drawing below. Draw a phasor
diagram labeled with the dierence in phase (in terms of ) between
adjacent phasors. What is now the smallest angle for which the
c) Now
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N
antennae which radiate equal intensity radio waves with a
wavelength are located as shown in the drawing below. What is now
the smallest angle for which the radio signal disappears? In the limit
that N becomes innite What is the smallest angle for which the
radio signal disappears?
c) Repeat part (a) for a diraction grating with 15,000 slits per centimeter.
SOLUTION: Now with 15,000 slits per centimeter, or d,1 = 1:5 106
slits/m, for = 434 nm, = 6:5 10,1 rad and for = 410 nm, =
6:15 10,1 rad.
SOLUTION: The signal disappears when the phasors add to zero; that is,
,1) . In the limit that N becomes innite,
when = 2N , and sin = (NNL
this becomes sin = L
7. Diraction Grating
a) A diraction grating with 2000 slits per centimeter is used to mea-
sure the wavelengths emitted by hydrogen gas. At what angles in the
rst-order spectrum would you expect to nd the two violet lines of wavelengths 434 and 410 nm?
SOLUTION: The rst order maximum for wavelength and slit spacing
d occurs at (for small angles) = =d. For this grating, d,1 = 2 105
slits/m so = 2 105 = m. For = 434 nm, = 8:6 10,2 rad and
for = 410 nm, = 8:2 10,2 rad.
b) With the grating described above, two other lines in the rst-order
hydrogen spectrum are found at angles 1 = 9:72 10,2 and 2 = 1:32 10,1. Find the wavelengths of these lines.
SOLUTION: Again using the relation = 2 105 = m, we get that at
1 = 9:72 10,2 , 1 = 486nm and at 2 = 1:32 10,1 , 2 = 660nm.
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