Unformatted text preview: University of Alabama
Department of Physics and Astronomy
Department of Electrical and Computer Engineering
PH 495/ECE 493 LeClair & Kung Spring 2011 Problem Set 5
Instructions:
1. Answer all questions below. All questions have equal weight.
2. Show your work for full credit.
3. All problems are due Friday 25 March 2011 by 11:59pm.
4. You may collaborate, but everyone must turn in their own work.
Problems 9 and 10 involve a takehome experiment, for which you will need 2 glass slides and a
planoconvex lens. These items will be provided in lecture on 10 March, or you may pick them up
from Dr. LeClair by appointment after that time.
1. Bekeﬁ & Barrett 8.2; Hecht 9.24 A radar antenna operating on a wavelength of 0.10 m is located
8 m above the water line of a torpedo boat. Treat the reﬂected beam from the water as originating
in a source 8 m below the water directly under the radar antenna. The dipole antenna is oriented
perpendicular to the plane of the page. 8m θ x R
(a) What is the altitude x of an airplane 12 km from the boat if it is to be in the ﬁrst interference
minimum of the radar signal?
(b) What is the total number of minima one observes as one scans the sky in the vertical plane as
a function of the angle θ, from θ = 0 to θ = π, keeping the distance R ﬁxed?
2. Bekeﬁ & Barrett 8.3 Two dipole radiators (e.g., the oscillating current segments we discussed
in class) are separated by a distance λ/2 along the x axis (halfwave dipole antenna). The dipoles
are oriented along z, as in the problem we worked in class. Assume the distance to the observation
point r satisﬁes r λ. (a) Plot the intensity of radiation in the x−y plane. Note the values of intensity at θ = 0, π/3, π/2, π
if the oscillators are in phase.i
(b) Repeat (a) if the oscillators are 180◦ out of phase.
(c) The oscillators are now spaced by a distance λ/4 and are 90◦ out of phase. Repeat (a). Note
that this conﬁguration would be very useful for a broadcast station in a coastal city, for example
...
3. Bekeﬁ & Barrett 8.5 We desire to superpose the oscillations of several simple harmonic oscillators
having the same frequency ω and amplitude A, but diﬀering from one another by constant phase
increments α; that is,
E(t) = A cos ωt + A cos (ωt + α) + A cos (ωt + 2α) + A cos (ωt + 3α) + · · · (1) (a) Using graphical phasor addition, ﬁnd E(t); that is, writing E(t) = Ao cos (ωt + ϕ), ﬁnd Ao and
ϕ for the case when there are ﬁve oscillators with A = 3 units and α = π/9 radians.
(b) Study the polygon you obtained in part (a) and, using purely geometrical considerations, show
that for N oscillators
sin Nα/2
N−1
E(t) = (NA)
cos ωt +
α
N sin α/2
2 (2) (c) Sketch the amplitude of E(t) as a function of α.
The above calculation is the basis of ﬁnding radiation from antenna arrays and diﬀraction gratings.
4. Hecht 9.10 White light falling on two narrow slits emerges and is observed on a distant screen.
If red light (λo = 780 nm) in the ﬁrstorder fringe overlaps violet in the secondorder fringe, what is
the latter’s wavelength?
5. Hecht 9.26 A soap ﬁlm surrounded by air has an index of refraction of 1.34. if a region of the
ﬁlm appears as bright red (λo = 633 nm) in normally reﬂected light, what is its minimum thickness
there?
6. Hecht 9.36 One of the mirrors in a Michelson interferometer is moved, and 1000 fringe pairs
i
You want to make a polar plot with intensity as the radial distance and reference the angle from the midpoint
between the two sources. Wolfram alpha is handy for this, http://wolframalpha.com.Try a query like “plot of r =
4cos^2((pi/4)*sin(theta) + pi/4)” shift past the hairline in a viewing telescope during the process. If the device is illuminated with
500 nm light, how far was the mirror moved?
7. Hecht 9.47 A glass camera lens with an index of refraction of 1.55 is to be coated with a cryolite
ﬁlm (n ≈ 1.30) to decrease the reﬂection of normally incident green light (λo = 500 nm). What
thickness should be deposited on the lens?
8. Bekeﬁ & Barrett 8.9 A plane electromagnetic wave of wavelength λo is incident on two long,
narrow slits, each having width 2a and separated by a distance 2b, with b a. One of the slits is
covered by a thin dielectric plate of thickness d, and dielectric coeﬃcient κ, with d chosen so that
√
( κ − 1)d/λo = 5/2.
The interference pattern due to the slits is observed in a plane a distance L from the slits, where L
is large enough so that the far ﬁeld approximations may be used, that is, the pattern depends only
on the angle θ from the normal to the slits, as shown.
(a) Consider eﬀects due to interference only. What is the condition for a maximum in the pattern?
Sketch the interference pattern. 2a
θ 2b
2a dielectric slab (b) Now include eﬀects due to both interference and diﬀraction. How is the intensity distribution
modiﬁed from that obtained in (a)? Let b/a = 10, sketch the resulting interferencediﬀraction
pattern. (Assume that all angles involved are small enough so that cos θ ≈ 1, and hence that the
optical path through the dielectric is independent of angle.)
For questions 9 and 10, download the procedure for the takehome experiment here:ii
ii These experiments have been adapted from Mavalvala, Nergis, Walter Lewin, and Wolfgang Ketterle, “8.03 http://ocw.mit.edu/courses/physics/803physicsiiivibrationsandwavesfall2004/
assignments/take_home_exp6.pdf
In these experiments, you will investigate two interference eﬀects. The phenomena are not hard
to observe, and the mathematical derivations and descriptions (found in your text) are not difﬁcult. However, until you see these eﬀects yourself and explore them a bit, you will not have a
true understanding of interference. Plus, the interference patterns can be quite striking and colorful!
9. Perform the interference experiment with two glass slides. Take care in making sure your slides
are clean and dustfree. Observing the fringes may be easier using a desk lamp close to the slides,
position yourself such that you see the mirrorlike reﬂection from the glass surface. If you do not
see the fringes clearly, try gently pressing the slides together with a pencil so they make better
contact. For a color ﬁlter, any colored glass or plastic will do (perhaps a bottle). If you can’t ﬁnd
anything, try using a colored light source, such as bright LED. An interesting variation of the fringe
patterns can be seen if you press (carefully!) on the slides with a pencil to slightly bow the slides
 you are essentially observing the strain ﬁeld created in the glass.
Brieﬂy, in a paragraph or two, discuss your ﬁndings qualitatively, and make an orderofmagnitude
estimate of the amount by which your slides deviate from being perfectly ﬂat.
10. Perform the experiment on Newton’s rings using your planoconvex lens and one of the glass
slides. Note that the rings are quite small and harder to observe than the fringes in the previous
experiment – using a second lens or magnifying glass to more easily observe the rings may be a
good idea.
Brieﬂy discuss your ﬁndings qualitatively. Note in particular why the rings have the size that they
do, and how you might make them bigger. Use the rings to roughly estimate the radius of curvature
of your lens. From an earlier lab, you should know how to verify the radius of curvature from the
focal distance of the lens. Physics III: Vibrations and Waves,” Fall 2004. (Massachusetts Institute of Technology: MIT OpenCourseWare),
http://ocw.mit.edu (Accessed 09 Mar, 2011). License: Creative Commons BYNCSA. University of Alabama
Department of Physics and Astronomy
Department of Electrical and Computer Engineering
PH 495/ECE 493 LeClair & Kung Spring 2011 Problem Set 5: Solutions
1. Bekeﬁ & Barrett 8.2; Hecht 9.24 A radar antenna operating on a wavelength of 0.10 m is located
8 m above the water line of a torpedo boat. Treat the reﬂected beam from the water as originating
in a source 8 m below the water directly under the radar antenna. The dipole antenna is oriented
perpendicular to the plane of the page. is there a m 8m θ x R
(a) What is the altitude x of an airplane 12 km from the boat if it is to be in the ﬁrst interference
minimum of the radar signal?
(b) What is the total number of minima one observes as one scans the sky in the vertical plane as
a function of the angle θ, from θ = 0 to θ = π, keeping the distance R ﬁxed?
Solution: From the geometry of the ﬁgure, referencing the crudely modiﬁed ﬁgure below, you
should be able to convince yourself that
(1) d2 = (x − h)2 + R2
2 d xh = (x + h) + R d h
h 2 θ θ
θ d’
R (2) 2 x
h The path diﬀerence between the direct ray (d) and the reﬂected ray d is thus
2
2−
(x − h)2 + R2
d − d = (x + h) + R
(3) The phase diﬀerence δ between the two rays is this path diﬀerence multiplied by k = 2π/λ plus an
extra phase shift of π for the reﬂected ray:
δ=
The intensity is then easily found:
2π
d −d +π
λ (4)
π
π
δ
I = 4Io cos
= 4Io cos2
d − d + π = 4Io sin2
d − d
2
λ
λ
2 (5) The minima in intensity occur when
δ
π
=
d − d = nπ
2
λ
nλ = d − d = (x + h)2 + R2 − (6)
(x − h)2 + R2 (7) The problem is then the “simple” matter of solving the above for x. It can be done, with some
tedium. First square both sides and isolate the remaining radical.
nλ = 2 (x + h) + R2 − (x − h)2 + R2
n2 λ2 = (x + h)2 + R2 + (x − h)2 + R2 − 2 (x + h)2 (x − h)2 + R4 + R2 (x + h)2 + (x − h)2
Square both sides and collect terms . . .
2
2
2
22
2 x + h + R − n λ = 2 (x2 − h2 )2 + R4 + 2R2 (x2 + h2 )
2
4 x2 + h2 + R2 + n4 λ4 − 4n2 λ2 x2 + h2 + R2 = 4 x4 + h4 − 2x2 h2 + R4 + 2R2 x2 + h2 4x4 + 4h4 + 4R4 − 8x2 h2 + 8x2 R2 + 8h2 R2 = 4x4 + 4h4 + 4R4 + 8x2 h2 + 8x2 R2
+ 2h2 R2 + n4 λ4 − 4n2 λ2 x2 + h2 + R2
16x2 h2 − 4n2 λ2 x2 = 4n2 λ2 h2 + R2 − n4 λ4 = n2 λ2 4h2 + 4R2 − n2 λ2
n2 λ2 4h2 + 4R2 − n2 λ2
2
x=
16h2 − 4n2 λ2
4R2 + 4h2 − λ2 n2
x = nλ
≈ 75.0 m
(8)
16h2 − 4λ2 n2 If we ﬁrst use the very reasonable approximation λ {R, h} and then the nearly as reasonable h R,
we can come to something a bit simpler: x = nλ √
4R2 + 4h2 − λ2 n2
R2 + h2
R
≈ nλ
≈ nλ
16h2 − 4λ2 n2
2h
2h (9) √
These latter two forms are readily derived by starting from Eq. 7, factoring out R2 + x2 + h2
from both terms, and approximating the resulting radicals for 2xh R2 + x2 + h2 using a binomial
expansion (since x is at minimum h, this is ok). Using the numbers given, one ﬁnds the ﬁrst
minimum at x ≈ 75.0 m.
How many minima must there be over the interval [0, π]? We have solved the problem of ﬁnding
the minima only over the interval [0, π/2]. However, owing to the symmetry of the problem, we
know that the interval [π/2, π] has the same number of minima.
Unfortunately, we have no obvious restriction on magnitude x from the information given, and
everything else is ﬁxed. We do know that at θ = 0, x = h, but as θ approaches π/2, x increases
without bound. What to do? The exact expression above does have an interesting condition built
in, however: x must be real. For this to be true, the numerator and denominator in Eq. 8 must
both be positive. That is,
2 2
R + h2
λ
2h
=
λ 4R2 + 4h2 − λ2 n2 > 0 =⇒ nmax = (10) 16h2 − 4λ2 n2 > 0 =⇒ nmax (11) Given that R h, it is clear that the second expression is more restrictive. Using the numbers
given, one ﬁnds n = 160. This is for the interval [0, π/2], the problem we have solved. The interval
(π/2, π] thus contains 159 minima (not double counting the minima at π/2, so in total over θ ∈ [0, π]
we have n = 319 minima. Alternate method The above is an exact method, though far from obvious. Given that both λ and h are tiny compared
to R, we need not solve the problem exactly. We can use the same dipole approximation we employed
previously to make things far simpler. The gist of the idea is that the source is very distant compared
to the spacing of the real and virtual sources (h << R), and the geometry looks more like this: A Δ
h x
r γ
θ θ R We assume the source is suﬃciently distant that the rays approaching the source can be approximated as parallel. Construct a line running from the intersection of the reﬂected ray with the water
to the direct ray, such that this line meets the direct ray at a right angle (point A in the ﬁgure).
The parallel ray approximation means that from point A rightward, both direct and reﬂected rays
traverse the same distance. The only geometrical phase diﬀerence is thus due to the path diﬀerence:
the reﬂected ray travels a distance r, while the direct ray travels a distance ∆. If we can ﬁnd the
diﬀerence r − ∆, we are nearly done.
From the ﬁgure above, the surface of the water is made up from angles θ, γ, and 90, such that
γ = 180 − 90 − 2θ = 90 − 2θ (12) h
sin θ
∆ = r sin γ (13) You can convince yourself that
r= (14) The path diﬀerence is then
h
h
r−∆=
− r sin γ =
−
sin θ
sin θ h
sin θ sin (90 − 2θ) = h
(1 − cos 2θ)
sin θ (15) Accounting for the phase shift on reﬂection, our total phase shift is then
δ=
2π
2π h
2π h
4πh
(r − ∆) + π =
(1 − cos 2θ) + π =
2 sin2 θ + π =
sin θ + π
λ
λ sin θ
λ sin θ
λ (16) The intensity is then
δ
π
2 2πh
2 2πh
I = 4Io cos
= 4Io cos
sin θ +
= 4Io sin
sin θ
2
λ
2
λ
2 (17) The condition for a minimum is 2πh
sin θ = nπ
λ or sin θ = nλ
2h (18) For a very distant source, the separation of the parallel rays is small compared to x, and we may approximate x ≈ R tan θ ≈ R sin θ = Rnλ/2h, which gives x ≈ 75 m in agreement with our previous result.
As for the number of minima, we need only note from the above expression that sin θ returns values
only in the interval [0, 1] over θ ∈ [0, π/2]. That is, since sin θ is at most 1, nλ
= sin θ 1
2h or n 2h
= 160
λ (19) This is precisely the condition we derived earlier; accounting for the angles [π/2, π], we end up with
a total of 2 × 160 − 1 = 319 minima.
Whether you prefer the ﬁrst or second method is really a matter of taste. I preferred the ﬁrst since
it gives an exact answer, though such precision is of limited utility  it is probably not worth the
extra diﬃculty. The advantage of the second method is that it is the same problem we’ve solved
many times already (it is just the double slit again, or Lloyd’s mirror) with a couple of small twists.
2. Bekeﬁ & Barrett 8.3 Two dipole radiators (e.g., the oscillating current segments we discussed
in class) are separated by a distance λ/2 along the x axis (halfwave dipole antenna). The dipoles
are oriented along z, as in the problem we worked in class. Assume the distance to the observation
point r satisﬁes r λ.
(a) Plot the intensity of radiation in the x−y plane. Note the values of intensity at θ = 0, π/3, π/2, π
if the oscillators are in phase.i
(b) Repeat (a) if the oscillators are 180◦ out of phase.
(c) The oscillators are now spaced by a distance λ/4 and are 90◦ out of phase. Repeat (a). Note
that this conﬁguration would be very useful for a broadcast station in a coastal city, for example
...
Solution: The intensity is
I = 4Io cos 2
πd
1
1
2π
sin ψ − (ϕ1 − ϕ2 ) = 4Io cos
sin ψ − (ϕ1 − ϕ2 )
λ
2
2
2 (20) with d = λ . For (a) and (b), we need only make polar plots with I/Io as the radial coordinate and
2
ψ as the angular coordinate. For (a) we have ϕ1 − ϕ2 = 0, and for (b) we have ϕ1 − ϕ2 = π. For
(c), the spacing is now λ/4 and ϕ1 − ϕ2 = π/4. Can you see now why this would be useful for a
transmitter in a coastal city? You want to make a polar plot with intensity as the radial distance and reference the angle from the midpoint
between the two sources. Wolfram alpha is handy for this, http://wolframalpha.com.Try a query like “plot of r =
4cos^2((pi/4)*sin(theta) + pi/4)”
i 4 2 3 2 1
1
0.5π 0.5π
0.75π 0.75π 0.25π 0.25π
π
4 π
4 3 2 1 0 1 2 3 3 2 1 0
1.25π 4 1 2 3 4 1.75π
1.5π
1 1.25π 1.75π
1.5π
1 2 3 2
4 Figure 1: (left) Intensity pattern for two oscillators with zero phase oﬀset, spaced at d = λ/2. The oscillators are at
y = ±λ/4. (right) The same oscillators with a phase diﬀerence of ϕ1 − ϕ2 = π. The radiation pattern is rotated by 90◦ . 4 3 2 1 0.5π
0.75π 0.25π π
4 3 2 1 0
1.25π 1 2 3 4 1.75π
1.5π
1 Figure 2: Intensity pattern for two oscillators with phase oﬀset ϕ1 − ϕ2 = π/2, spaced at d = λ/4. The oscillators are at
y = ±λ/8. 3. Bekeﬁ & Barrett 8.5 We desire to superpose the oscillations of several simple harmonic oscillators
having the same frequency ω and amplitude A, but diﬀering from one another by constant phase
increments α; that is,
E(t) = A cos ωt + A cos (ωt + α) + A cos (ωt + 2α) + A cos (ωt + 3α) + · · · (21) (a) Using graphical phasor addition, ﬁnd E(t); that is, writing E(t) = Ao cos (ωt + ϕ), ﬁnd Ao and
ϕ for the case when there are ﬁve oscillators with A = 3 units and α = π/9 radians.
(b) Study the polygon you obtained in part (a) and, using purely geometrical considerations, show
that for N oscillators
E(t) = (NA)
sin Nα/2
N−1
cos ωt +
α
N sin α/2
2 (22) (c) Sketch the amplitude of E(t) as a function of α. The above calculation is the basis of ﬁnding radiation from antenna arrays and diﬀraction gratings.
Solution: You can ﬁnd a quick solution using a phasor diagram here:
http://ocw.mit.edu/courses/physics/803physicsiiivibrationsandwavesfall2004/
assignments/soln10.pdf
We will solve the N oscillator problem by more direct means. Each oscillator has a phase oﬀset of
α from its neighbor, so the Nth oscillator has a phase oﬀset of (N − 1) α from the ﬁrst. Since the
total ﬁeld from all N oscillators is the sum of their individual electric ﬁelds, and thusii
The symbol is the operator that takes the real part of expression, and is the operator that takes the
an
imaginary part. Also recall the sum of a ﬁnite geometric series n−1 ark = a(1 − rn )/(1 − r).
0
ii Etot = N−1
A cos (ωt + nα) = A n=0 N−1
cos (ωt) cos (nα) − sin (ωt) sin (nα) n=0 = A cos (ωt) N−1
cos (nα) − A sin (ωt) n=0 N−1
(23)
(24) sin (ωt) cos (nα) n=0 −
−
cos N2 1 α sin Nα
sin N2 1 α sin Nα
2
2
= A cos ωt
− A sin ωt
sin α
sin α
2
2
sin Nα
(N − 1) α
(N − 1) α
2
=A
cos (ωt) cos
− sin (ωt) sin
sin α
2
2
2
Nα
sin 2
(N − 1) α
sin Nα/2
N−1
(NA)
=A
cos ωt +
=
cos ωt +
α
sin α
2
N sin α/2
2
2 (25)
(26)
(27) Thus, total ﬁeld is that of a single oscillator of amplitude and phase
sin Nα
2
Ao = A
sin α
2
(N − 1) α
ϕ=
2 (28)
(29) One can also perform the sum using complex exponentials, which makes it a simple geometric
series. Given N = 5, A = 3 and α = π/9, one ﬁnds Ao ≈ 13.23 and ϕ = 2π/9 = 40◦ . In order to sketch
the amplitude as a function of α, we should recall that
lim β →0 sin Nβ
=N
β (30) which means that the intensity has zeros at Nα/2 = nπ unless sin (α/2) = 0, in which case we have
a maximum. Below are plots of I/NA versus α for N = 5 (red) and N = 6 (black) for ωt = 0. Note
the primary maxima at α = 2nπ and the (N − 2) secondary maxima in between.
1 0.75 0.5 0.25 10 7.5 5 2.5 0 2.5 5 7.5 10 4. Hecht 9.10 White light falling on two narrow slits emerges and is observed on a distant screen.
If red light (λo = 780 nm) in the ﬁrstorder fringe overlaps violet in the secondorder fringe, what is
the latter’s wavelength?
Solution: The interference condition for a fringe to occur is
sin θm ≈ θm = mλ
a (31) where a is the slit spacing. If θred,m=1 = violet,m=2 , this implies
1
λviolet = λred = 390 nm
2 (32) 5. Hecht 9.26 A soap ﬁlm surrounded by air has an index of refraction of 1.34. if a region of the
ﬁlm appears as bright red (λo = 633 nm) in normally reﬂected light, what is its minimum thickness
there?
Solution: If the ﬁlm appears red, it is because the thickness is such that interference creates a
minimum in intensity for red light at wavelength λo in vacuum. These minima occur for a ﬁlm of
thickness d and index n at
d cos θt = (2m + 1) λf
4 (33) where the wavelength of light in the ﬁlm λf relates to the vacuum wavelength via λf = λo /n. If we
consider the intensity at normal incidence (θt = 0) and the minimum thickness (corresponding to
m = 0),
d= λf
λo
=
≈ 118 nm
4
4n (34) 6. Hecht 9.36 One of the mirrors in a Michelson interferometer is moved, and 1000 fringe pairs
shift past the hairline in a viewing telescope during the process. If the device is illuminated with
500 nm light, how far was the mirror moved?
Solution: A fringe pair will shift pass the hairline whenever one of the mirrors moves by λ/2. If
N fringe pairs move past the hairline, one of the mirrors must have moved by a distance
1
∆d = Nλ
2 (35) With N = 1000 and λo = 500 nm, ∆d = 250 µm.
7. Hecht 9.47 A glass camera lens with an index of refraction of 1.55 is to be coated with a cryolite ﬁlm (n ≈ 1.30) to decrease the reﬂection of normally incident green light (λo = 500 nm). What
thickness should be deposited on the lens?
Solution: Reﬂectivity minima occur for a ﬁlm of thickness d and index n when
d cos θt = (2m + 1) λf
4 (36) where the wavelength of light in the ﬁlm λf relates to the vacuum wavelength λo via λf = λo /n. If
we consider the intensity at normal incidence (θt = 0), the minimum thickness must be
d = (2m + 1) λf
λf
λo
=
=
≈ 96 nm
4
4
4n (37) 8. Bekeﬁ & Barrett 8.9 A plane electromagnetic wave of wavelength λo is incident on two long,
narrow slits, each having width 2a and separated by a distance 2b, with b a. One of the slits is
covered by a thin dielectric plate of thickness d, and dielectric coeﬃcient κ, with d chosen so that
√
( κ − 1)d/λo = 5/2.
The interference pattern due to the slits is observed in a plane a distance L from the slits, where L
is large enough so that the far ﬁeld approximations may be used, that is, the pattern depends only
on the angle θ from the normal to the slits, as shown.
(a) Consider eﬀects due to interference only. What is the condition for a maximum in the pattern?
Sketch the interference pattern. 2a
θ 2b
2a dielectric slab (b) Now include eﬀects due to both interference and diﬀraction. How is the intensity distribution
modiﬁed from that obtained in (a)? Let b/a = 10, sketch the resulting interferencediﬀraction
pattern. (Assume that all angles involved are small enough so that cos θ ≈ 1, and hence that the
optical path through the dielectric is independent of angle.)
Solution: Consider the revised ﬁgure below, which places a screen a distance L {a, b} from the
slits. An observation point on the screen is then a distance r1 from one slit and and r2 from the
other, as well as a distance R from the midpoint of the two slits. L
r1 2a R
θ 2b r2
2a
dielectric slab The law of cosines allows us to ﬁnd the distances r1 and r2 in terms of b, R and θ
r2 = b2 + R2 − 2bR cos (θ + 90) = b2 + R2 + 2bR sin (θ)
2 (38) r2 = b2 + R2 − 2bR sin (θ)
1 (39) The diﬀerence of the squared diﬀerences is then
r2 − r2 = 4bR sin (θ) = (r1 + r2 ) (r2 − r1 )
2
1
r2 − r1 = 4bR sin (θ)
r1 + r2 (40)
(41) One can solve this analytically for r2 in terms of r1 and then compute r2 − r1 exactly.iii However,
since we are explicitly encouraged to use the far ﬁeld approximation, we may as well save a bit
of tedium. If we make the approximation that θ is small, amounting to the condition L ≈ R, then
iii This results in r2−r1 ≈ 2r2 − 8bR sin θ, which does not depend only on θ as the far ﬁeld approximation requires.
2
r1 ≈ r2 and thus r1 + r2 ≈ 2R. We cannot do the same for r2 − r1 , however, since we are interested in
diﬀerence between r2 and r1 on the scale of the incident wavelength. This leads us to
r2 − r1 ≈ 4bR sin (θ)
2R (42) This is the geometric path diﬀerence between the two beams, exactly as we have previously derived
for the two slit problem. Ignoring the ﬁnite width of the slits for the moment, this would be
suﬃcient to calculate the two slit interference pattern. In the present case, we must also take into
account the propagation delay suﬀered by the second beam resulting from its transit through the
dielectric. In crossing a distance d through the dielectric, the second beam requires a time
t2 = d
dn
=
v
c (43) while the ﬁrst transiting only vacuum through the same distance d requires a time
t1 = d
c (44) The second beam therefore suﬀers a delay of
δt2 =
d
d √
(n − 1) =
κ−1
c
c (45) √
where we have used the result n = κ in a medium where µ ≈ µo . This time delay implies a phase
delay for the second beam relative to the ﬁrst of
d √
ωd √
κ−1 =
κ − 1 = 5π
(46)
c
λ
√
Here we have used λ = c/f = 2πc/ω and the given relationship ( κ − 1)d/λo = 5/2. We must add
this phase delay to the geometric phase delay k (r2 − r1 ) to ﬁnd the total phase diﬀerence between
the two beams:
ωδt2 = (47) δtot = 2kb sin θ + 5π
This gives the intensity on the screen asiv
I = 4Io cos 2 δtot
2
5π
= 4Io cos kb sin θ −
2
2 (48) We will have maxima when the argument of sin2 is an integer multiple of π:
5π
kb sin θ −
= mπ
2 or λ
2b sin θ =
2
5
m+
2 (49) Note that whether we pick +5π/2 or −5π/2 makes no diﬀerence here, it physically only amounts to asking whether
we are considering points above or below the midpoint of the screen, and ultimately makes no diﬀerence at all.
iv Since m is just an integer, what this formula is really telling us is that the path diﬀerence 2b sin θ
must be a halfinteger multiple of λ/2. We could just as well reindex the starting point of m and
write
λ
r2 − r1 ≈ 2b sin θ =
2 1
m+
2 (50) The gist of the matter is that the interference pattern now has a minimum at θ = 0 as a result of
the dielectric slab, rather than a maximum. Otherwise, it looks just like the usual twoslit problem.
The inclusion of diﬀraction essentially turns each slit into an array of point sources rather than
a single point source. When we take into account the ﬁnite slit width, in the farﬁeld regime we
can imagine each point within a slit acts as a new source of spherical waves which will interfere
with each other. We can model this as an array of point sources, as in problem 3, letting N → ∞.
Moreover, now the constant phase diﬀerence between neighboring oscillators makes sense  since
each successive evenlyspaced source is slightly farther from the observation point, this phase oﬀset
is accounting for the propagation delay due to the spacing between oscillators.
If we have a slit of width 2a made up of N oscillators, then the spacing between oscillators is
d = 2a/N, or N = 2a/d. Our slit is the case that d → 0 with ﬁxed 2a. From problem 3, the total
ﬁeld of N oscillators in an array, each of intensity A, was
sin Nα/2
N−1
E = (NA)
cos ωt +
α
N sin α/2
2 (51) The phase diﬀerence α for oscillators spaced a distance d from each other can be written (in analogy
to our work above) α = kd sin ψ, where ψ ≈ θ is the angle to the point of interest. Adding 2a = Nd
and squaring to ﬁnd the intensity (and noting k = 2π/λ),
sin2 d 2πa sin ψ
λ
I=A
2 2πa
sin
sin ψ
λN (52) 2 Taking the limit that N → ∞, with ψ ≈ θ and deﬁning Io ≡ A2 , the intensity of a slit of ﬁnite width
2a is
2 I = N Io sin β
β 2 with β= 2πa
sin θ
λ (53) In the end, accounting for the ﬁnite width of the slits merely modulates the intensity by a factor
sin2 (β)/β. Thus, for the problem at hand,
I = 4Io sin 2πa λ sin θ
2πa
λ sin θ 2 cos 2 2πb
5π
sin θ −
λ
2 (54) The intensity plot is left as an exercise to the reader . . . ...
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This note was uploaded on 11/22/2011 for the course ECE 350 taught by Professor Staff during the Spring '08 term at Alabama.
 Spring '08
 Staff

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