Answers  Physical Optics


1.

 The concentric circles are crosssections of spherical
wave fronts that pass through the crests of the waves. The
distance between crests equals one wavelength λ.
 The distance SP is from one crest to another or λ.
 The distance SP’ is 2 λ.
 The distance SP” is 5 λ.
 The distance between a crest and a trough is λ/2.
The wavefronts through troughs would be spheres halfway
between the spheres that represent the wavefronts through
the crests. In the figure you would represent their crosssections
with circles.


2.

 Counting the number of crests from S_{1 }to P
you find S_{1}P = 8 λ.
 Counting the number of crests from S_{2}P = 7 λ.
 The path difference S_{1}P  S_{2}P =
λ.
 For a path difference of an integral number of wavelengths,
there will be constructive interference.


3.

Frequency f is the number
of oscillations of the source of the wave per second, wavelength
λ is the distance traveled
by the wave in one complete oscillation of the source and velocity
is the distance traveled by the wave in one second.
number
of oscillations 
x 
distance
traveled 
= 
distance
traveled 
second 
oscillation 
second 
f 
x 
λ 
= 
v 


4.

For the first minimum, the path difference
= λ/2 = (10  8)ft = 2
ft and the wavelength λ=
4 ft. The speed of sound v = λf
= (4 ft)(275 s^{1}) = 11 x 10^{2 }ft/s.


5.

 For a maximum the path difference equals an integral number
of wavelengths, λ, 2 λ,
3 λ, . . . S_{1}P
= m λ, where m = 0, 1,
2, 3, . . . .
 In triangle BS_{1}S_{2 }of Fig. 3 above,
sin Θ = S_{1}B/d
= m λ/d,
or m λ = d sin Θ
for a maximum.
 For a minimum the path difference equals λ/2,
3 λ/2, 5 λ/2,
. . .
S_{1}P = (m + 1/2) λ,
where m = 0, 1, 2, 3, . . . .
In triangle BS_{1}S_{2}, sin Θ
= S_{1}B/a = (m + 1/2) λ/d,
or m(1 + 1/2) λ = d sin
Θ for a minimum.


6.

For Fig. 4b above,
 The resultant amplitude A = A_{1 }+ A_{2}
= 2A_{o}, where A_{1} = A_{2}
= A_{o}.
 The phase difference = 0, 2 π,
3 π, . . . or ΔΦ
= 2m π,
m = 0, 1, 2, 3, . . .
For Fig. 4c above,
 The resultant amplitude A = A_{1 } A_{2}
= 0, where A_{1} = A_{2} = A_{o}.
 The phase difference = π,
3 π, 5 π,
. . . or ΔΦ =
2 π(m + 1/2),
m = 0, 1, 2, 3, . . .


7.

ΔΦ =
2 π(r_{1}  r_{2})/ λ
= 2 π (path difference)/ λ.
 For constructive interference, ΔΦ=
2 πm = 2 π
(r_{1}  r_{2})/ λ
or path difference = (r_{1}  r_{2}) = m λ,
m = 0, 1, 2, 3, . . .
 For destructive interference, ΔΦ=
2 π (1 + 1/2)m = 2 π
(r_{1}  r_{2})/ λ
or path difference = (r_{1}  r_{2}) = m(1
+ 1/2) λ, m = 0,
1, 2, 3, . . .


8.

In Fig. 5 above, sin Θ=
y_{m}/(L^{2} + y_{m}^{2})^{1/2
}≈ y_{m}/L^{ }= tan Θfor small Θ.
 For constructive interference sin Θ
= m λ/d ≈ y_{m}/L.
Dropping the approximate sign, y_{m }= m λL/d.
 Δy = y_{m+1}
 y_{m }= (m + 1) λL/d
 m λL/a = λL/d.
 For destructive interference sin Θ
= (m + 1/2) λ/d ≈
y_{m}/L.
 y_{m }= (m + 1/2) λL/d.
 (b) Δy = y_{m+1}
 y_{m }= (m + 1 + 1/2) λL/d
 (m + 1/2) λL/d
= λL/d.


9.

From Fig. 4a below, A_{x}
= A_{2} + A_{1 }cos ΔΦ.
A_{y }= A_{1 }sin ΔΦ.
 A^{2} = [(A_{2} + A_{1 }cos ΔΦ)^{2}
+ (A_{1 }sin ΔΦ)^{2}]
A^{2} = A_{2}^{2} + 2A_{2}A_{1}
cos ΔΦ + A_{1}^{2}
cos^{2 }ΔΦ+
A_{1}^{2} sin^{2} ΔΦ
A^{2} = A_{2}^{2} + 2A_{2}A_{1}
cos ΔΦ + A_{1}^{2}
= 2A_{o}^{2}(1 + cos ΔΦ)
A^{2} = 2A_{o}^{2}(2 cos^{2 }ΔΦ/2)
A^{2} = 4A_{o}^{2} cos^{2 }ΔΦ/2.
 I/I_{o} = (A/A_{o})^{2 }= 4 cos^{2
}ΔΦ/2. I
= I_{o }cos^{2 }ΔΦ/2,
where ΔΦ/2= πd
sin Θ/ λ.


10.

 For direction (a) the waves from each antenna travel the
same distance to a distant point. The path difference and
the phase difference are zero.
A = 3A_{o} and I = 9I_{o}. Fig.
for 10(a)
 For direction (b) the path difference between the wave
from antenna 2 and 3 is 90 m sin 30^{o} = 45 m =
λ/4.
ΔΦ = 2 π(path
difference)/ λ= 2 π( λ/4)/ λ=
π/2. The path difference
between antenna 1 and 3 is 180 m sin 30^{o} = 90
m = λ/2.
ΔΦ= π.
A = A_{2 }= A_{o}. I = I_{o}. Fig.
for 10(b)
 For direction (c), the path difference between the wave
from antenna 2 and 3 is 90 m = λ/2.
ΔΦ =2 π(path
difference)/ λ= 2 π( λ/2)/ λ=
π. The path difference
between antenna 1 and 3 is 180 m = λ.
ΔΦ= 2 π.
A = A_{3 } A_{2 }+ A_{1 }=
A_{o}. I = I_{o}. Fig.
for 10(c)


11.

When the unpolarized light goes through the first polarizer,
its irradiance drops by a factor of two. With I_{o
}= 1000 W/m^{2}, I_{1} = 500 W/m^{2}.
When it passes through the second polaroid, I_{2}
= I_{1} cos^{2} 60^{o }= 500 W/m^{2}
(1/4) = 125 W/m^{2}.


12.

 For constructive interference m λ
= d sin Θ_{m},
where m is an integer
= 0, 1, 2, 3 . . . ., λ
is the wavelength of the light, d is the distance between
the sources or slits, and Θ_{m
}is the angle for the mth maximum. Thus sin Θ_{3}
= m λ/d = 3(656 x 10^{9
}m)/(1.57 x 10^{5} m) = 0.125. Θ_{3} = 7.2^{o}.
 For small angles sin Θ
≈ tan Θ ≈
y/L.
Thus y_{3} ≈ LΘ_{3}
= 2.00 m(0.125) = 0.25 m.


13.

In general, λ_{o}
= c/f, where f is the frequency of the light and λ_{o
}and c are the wavelength and the speed of light in a
vacuum or approximately in air, respectively. The frequency
is a property of the source and not of the medium. When a
wave enters a different medium, its wavelength and speed change,
but its frequency remains the same. For a film with index
of refraction n, the wavelength λ_{n}
= v/f = (c/n)/f = (c/f)(1/n) = λ_{o}/n,
since n = c/v or
v = c/n.


14.

When a pulse goes from a less to denser
medium, the reflected wave experiences a 180^{o} or
π phase change. When it
goes from a more to less dense medium, there is no change
of phase for the reflected wave. (Fig. for #14). From a crest
to a trough is a distance of λ/2.
A change of phase of 180^{o} or π
is equivalent to a path difference of λ/2.
Notice that there is never a change of phase for the transmitted
ray at an interface. The speed of light is greatest in a vacuum
or approximately air with index of refraction n = 1 = c/v.
The higher the index of refraction, the smaller the speed
in the medium. When light goes from medium 1 to medium 2,
the reflected ray will experience a phase change of π if
n_{2 }> n_{1}. There will be no phase
change if n_{2} < n_{1}.


15.

y_{m }= m λL/d
or
m = y_{m}d/L λ=
(1.50 x 10^{3} m)(1.00 x 10^{3 }m)/(1.00
m)(500 x 10^{9} m) = 3.


16.

Δy
= λL/d.
The distance between maxima Δy
is directly proportional to the wavelength λand
inversely proportional to the distance d between the slits.
If the wavelength is halved, the distance between maxima is
halved. If the distance between the slits is doubled, the
distance between maxima is halved. The combination means the
distance between maxima is reduced by a factor of four.
Δy’ = Δy/4.


17.

The distance between maxima Δy
is directly proportional to the wavelength λ.
The wavelength of light in water λ_{n
} of index of refraction n equals the wavelength λ_{o}
in air or a vacuum divided by n. Since n > 1,
λ_{n }is less
than λ_{o }and
the distance between maxima is smaller when the interference
pattern is observed under water.


18.

The distance from A to B is 5.00000 cm. The path from A to
the mirror to B is 1.60010 cm + 3.40020 cm = 5.00030 cm.
The difference in path is (5.00030 cm + λ/2)
 5.00000 cm = 0.00030 cm + λ/2.
The addition of λ/2 to
the path from A to the mirror to B occurs because of the change
of phase of π when the
light is reflected from the mirror. For a minimum at B, the
path difference = 0.00030 cm + λ/2
= (m + 1/2) λ or
0.00030 cm = m λ.
For m = 1, λ= 0.00030
cm = 30 x 10^{7} m.
For m = 2, λ= 0.00015
cm = 15 x 10^{7 }m.


19.

When the light goes from air to oil, the reflected wave 1 experiences
a change of phase of π.
The light that enters the oil is partly reflected at the oilwater
interface. Again at this interface, the reflected wave experiences
a change of phase of π,
but it has no change in phase as it goes from the oil back into
the air as the transmitted ray 2. The change of phase are additive.
For both reflections, the total change of phase is π
+ π = 2 π,
which is equivalent to no change in phase.
There is, however a path difference as the ray travels down
and back up in the oil. For almost normal incidence we say the
thickness of the oil there is d and the path difference
2d = m λ_{f},
m = 0, 1, 2, 3 . . . for constructive interference,
or
2d = (m + 1/2) λ_{f},
m = 0, 1, 2, 3 . . . for destructive interference,
where λ_{f} is the wavelength of the
light in the film.
 For the outer or thinnest regions of the drop, d approaches
zero and there is no overall phase change or thickness so
there is a maximum intensity of light.
 For a maximum 2d = m λ_{f
}=_{ }m λ_{o}/n_{f}.
Thus possible d's are d_{o} = 0, d_{1}=
1(580 nm)/2(1.20) = 242 nm.
d_{2} = 2(580 nm)/(2)(1.20)= 484 nm, etc.


20.

Now there is a change of phase at the interface between the
air and the soap film so that ray 1 is reflected with a change
of phase of π. But
there is no change of phase at the second filmair interface
because now the ray is going from n_{f} to n, where
n_{f} > n. This is similar to adding (or subtracting)
a path difference of λ_{f}/2.
For constructive interference, the path difference = 2d = (m
+ 1/2) λ_{f} =
(m + 1/2) λ_{o}/n_{f}_{
}.
 For m = 2, a maximum occurs for d = (2 + 1/2)(580
nm)/2(1.34) = 541 nm.
A minimum occurs for a path difference = 2d = m λ_{f},
which would have been the condition for a maximum
had there been no change in phase of the reflected ray.
Now for a minimum, d = m λ_{f}/2
= m λ_{o}/2n_{f}.
 For m = 2, a minimum occurs for d = 2(580 nm)/2(1.34)
= 433 nm.


21.

The sound is diffracted as it goes through
the "slit" of the door with the longer wavelengths
"bent" the most around the "slit". These
wavelengths will be accentuated. The longer wavelengths correspond
to lowerfrequency notes.


22.

The wavelengths of sound are much greater
than the wavelengths of visible light. Thus it is much easier
to detect interference and diffraction properties of sound
waves.


23.

For small angles, the position of the minima y_{m’}
is directly proportional to the wavelength λ and
inversely proportional to the slit width a.
 When the wavelength is halved, but slit width remains
the same, the first minimum is closer to the central maximum
and the distance between minima is smaller (Fig. for #23a).
y_{1} for (a) is half the original y_{1}
and the distance between minima Δy
for (a) is half the original Δy.
 When the slit width is halved, but wavelength remains
the same, the first minimum is further from the central
maximum and the distance between minima is greater (Fig.
for #23b).
y_{1} for (b) is twice the original y_{1}
and the distance between minima Δy
for (b) is twice the original Δy.


24.

For a minimum sin Θ_{m’}
= m’ λ/a.
For m’ = 1, the width of the slit a = 1.00 x 10^{4}
m and
λ = 580 x 10^{9 }m,
sin Θ_{1 }= 1(580
x 10^{9} m)/10^{4} m = 0.00580 ≈ Θ_{1}.
The width of the central maximum ≈ 2LΘ_{1}
= 2(2.0m)(0.0058) = 2.32 x 10^{2} m. Plot of the intensity
as a function of angle is shown in the figure above.


25.

The angular limit of the resolution for the terrestrial
telescope =
Θ_{a} = 1.22 λ/D
= (1.22)(550 x 10^{9} m)/(6.5 x 10^{2}
m) = 1.03 x 10^{5}.


The minimum distance of separation at a distance of 1.00
km is
x = (10^{3} m)(1.03 x 10^{5}) = 1.03
x 10^{2} m = 1.03 cm.
(See Fig. for # 25, to the right) 


26.

In general for a maximum, sin Θ_{m
}= m λ/d.
In this problem, d = 1 divided by the number of lines
per meter
= 1/(2.00 x 10^{5}/m) = 5 x 10^{6 }m.
For λ= 653.4 x 10^{9
}m and m = 1,
sin Θ_{1} = (1)(653.4
x 10^{9 }m)/(5 x 10^{6} m) = 0.131 and Θ_{1}
= 7.51^{o}.
For λ= 653.4 x 10^{9
}m and m = 3,
sin Θ_{1} = (3)(653.4
x 10^{9 }m)/(5 x 10^{6} m) = 0.393 and Θ_{3}
= 23.1^{o}.
For λ= 580.8 x 10^{9
}m and m = 1,
sin Θ_{1} = (1)(580.8
x 10^{9 }m)/(5 x 10^{6} m) = 0.116 and Θ_{1}
= 6.67^{o}.
For λ= 580.8 x 10^{6
}m and m = 3,
sin Θ_{1} = (3)(580.8
x 10^{9 }m)/(5 x 10^{6} m) = 0.348 and Θ_{3}
= 20.4^{o}.
For the first order, the angular separation for the α
and β lines is
(7.51  6.67)^{o } = 0.84^{o}.
In the third order, the angular separation for the α
and β lines is
(23.1  20.4)^{o } = 2.7^{o}.
You get greater angular separation of the lines in the higher
orders.


27.

 ΔΦ = (2 πL/ λ)(n_{2}
n_{1}).
Light from S_{2} that passes through a medium of
index of refraction n_{2} = 1.5.
Light from S_{1} that passes through a medium of
index of refraction n_{1} = 1.0 of equal length
L = 0.50 x 10^{6} m.
The wavelength λ of
both sources is 500 x 10^{9 }m.
ΔΦ = (2 πL/ λ)(n_{2}
n_{1}) = [2 π(0.50
x 10^{6} m)/(500 x 10^{9 }m)](1.5  1.0)
= π.
 A phase difference of πor 180^{o} produces destructive interference.
 For a phase difference of 2 πand constructive interference
2 π = (2 πL'/500
x 10^{9 }m)(1.5  1.0) or
L' = 2L = 1.0 µm.


28.

I = I_{o} cos^{2
}Θ.
I/I_{o } = 1/2 = cos^{2} Θ.
cos^{ }Θ
= (1/2)^{1/2}.
cos^{1 }(1/2)^{1/2 }= 45^{o
}


29.


After going through
the first filter, the light has an amplitude of A_{o}
and its polarization is vertical. When this light goes
through the second filter that is at an angle of 45^{o
} with the axis of the first filter, the amplitude
is the projection of A_{o} on the new axis which
is A_{o} cos 45^{o}. The third
filter is at an angle of 90^{o } with the first
filter, but at an angle of 45^{o } with the second
filter. After passing through the third filter, the amplitude
is
A_{o} cos 45^{o}(cos 45^{o})
=
A_{o }cos^{2} 45^{o} =
A_{o }[(1/2)^{1/2}]^{2 }=
A_{o}/2.
The intensity is proportional to the amplitude square.
I/I_{o }= (A_{o}/2)^{2}/A_{o
} or
I = I_{o}/4 = 25% I_{o.} 


30.

Let n_{1} = 1 and
n_{2 }= 2.42.
tan Θ_{1 }= 2.42.
Θ_{1}
= 67.6^{o}.


31.

Since the polarizing angle found in #30
was 67.6^{o } it can't be diamond.
tan 62.5^{o} = 1.92, which is the index of refraction
of zircon.

