oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 4 2007, 11:00 pm
Question 1, chap 33, sect 3.
part 1 of 1
10 points
In a particular medium the speed of light
is three-fourths that of the speed of light in
vacuum and the electric field amplitude has a
maximum value of 4.4 mV/m.
The speed of light is 2.99792 × 108 m/s .
Calculate the maximum value of the magnetic field.
Correct answer: 1.95691 × 10−11 T (tolerance
± 1 %).
Explanation:
Let : E = 4.4 mV/m and
c = 2.99792 × 108 m/s .
c
The index of refraction is n ≡ , so the index
v
of refraction of light in this particular medium
is four-thirds.
According to Maxwell’s laws, for an electromagnetic wave, the relation between the
amplitude of the electric field and the magnitude of the magnetic field is given by
E=
c
B = vB,
n
where n is the index of refraction and v is
the velocity of light in that medium. For
3
4
since v = c , so the
our problem n =
3
4
maximum value of the magnetic field is
4E
3 c
V
4.4 mV/m
4
· 3
=
8
3 2.99792 × 10 m/s 10 mV
B=
= 1.95691 × 10−11 T .
Question 2, chap 33, sect 5.
part 1 of 1
10 points
Consider a monochromatic electromagnetic
plane wave propagating left to right (as shown
below). At a particular point in space, the
magnitude of the electric field has an instantaneous value of 535 V/m.
1
The permeability of free space is 4π ×
10−7 N/A2 , the permittivity of free space is
8.85419 × 10−12 C2 /N/m2 and the speed of
light is 2.99792 × 108 m/s.
propagation direction
E
B
What is the instantaneous magnitude of the
Poynting vector at the same point and time?
Correct answer: 759.761 W/m2 (tolerance ±
1 %).
Explanation:
Let :
E = 535 V/m ,
c = 2.99792 × 108 m/s ,
µ0 = 4π × 10−7 N/A2 .
and
Basic Concepts:
E
=c
B
b ×B
b =S
b.
E
~ is given by
The Poynting vector S
~ = 1 E
~ ×B
~.
S
µ0
Solution: For a plane, electromagnetic
~ and B
~ are always perpendicular to
wave, E
each other and to the direction of propagation
of the wave.
The magnetic field is
535 V/m
B=
2.99792 × 108 m/s
= 1.78457 × 10−6 T .
In this case, the Poynting vector is in the
direction of propagation and has magnitude
EB
S=
µ0
(535 V/m) (1.78457 × 10−6 T)
=
4π × 10−7 N/A2
= 759.761 W/m2 .
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 4 2007, 11:00 pm
2
Top View
Question 3, chap 33, sect 5.
part 1 of 1
10 points
blackened
0.7 cm
0.7 cm
torsion
balance
0.7 cm
blue
surface
mirrored surface
The plates are squares whose sides are of
length 0.7 cm. A point light source is then
placed at a distance 3 m away from the rod,
as shown below. The point light source radiates an average power of 25 kW. The torsion
balance keeps the rod from turning by counteracting any torque on the rod.
Assume: The direction of the EM waves are
nearly perpendicular to each plate’s surface
(i.e., a distance 3 m from each plate), the
light is radiated isotropically from the source,
and uniformly distributed over the surface of
each plate.
light
source
7 cm
Two plates, one black (perfectly absorbing)
and one mirrored (perfectly reflecting), are
affixed one at each end of a 7 cm rod, which
is then suspended at its center from a torsion
balance (see figure below).
mirrored
3m
Figure: Not drawn to scale.
What is the magnitude of the torque k~τ k
exerted on the torsion balance by the rod?
Correct answer: 1.26454 × 10−12 N · m2 (tolerance ± 1 %).
Explanation:
Let : c = 2.99792 × 108 m/s ,
ℓ = 7 cm ,
a = 0.7 cm ,
d = 3 m , and
Pav = 25 kW = 25000 W .
1 ~
~ deThe Poynting vector ~S =
E×B
µ0
scribes the rate of flow of energy in an EM
wave. Time average of S is equal to the intensity. In the present case, since ℓ ≪ d, intensity
at each of the mirrors is approximately equal
to the intensity at a distance d from the point
source. Hence
P
S=I=
4 π d2
25000 W
=
4 π (3 m)2
= 221.049 kg/s3 .
Thus the pressure p on the black plate is
p=
S
c
221.049 kg/s3
2.99792 × 108 m/s
= 7.3734 × 10−7 kg/m · s2 .
=
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 4 2007, 11:00 pm
The magnitude of the force is then
~ b k = (Pressure) (Area)
kF
= p a2
= (7.3734 × 10−7 kg/m · s2 )
× (0.007 m)2
= 3.61296 × 10−11 N .
A vertically pulsed laser fires a 1000 MW
pulse of 200 ns duration at a small 10 mg pellet at rest. The pulse hits the mass squarely
in the center of its bottom side.
The speed of light is 3 × 108 m/s and the
acceleration of gravity is 9.8 m/s2 .
h
y
The definition of torque is
~.
~τ = ~r × F
For both plates, the distance from the center
of the plate to the point of rotation is
ℓ
r= .
2
~ are perpendicuAlso, for both plates ~r and F
lar. For the black plate
k~τb k = r Fb .
For the mirrored plate
k~τm k = r Fm = 2 r Fb .
Using the right hand rule, we see that the
~ b will be pointing up, and the
torque due to F
~ m will be pointing down. So the
one due to F
total torque is
k~τ k = k~τb k − k~τm k
= r Fb − r Fm
= r Fb − 2 r Fb
= −r Fb
7 cm
1
=−
2 100cm/m
× (3.61296 × 10−11 N)
= −1.26454 × 10−12 N m
b
b
b
b
b b b
b b
b
b
b
Question 4, chap 33, sect 5.
part 1 of 1
10 points
b
b
T is the time
to reach its
b
maximum
b
height
b
b
h
b
v0
b
b
b
b
b
10 mg
t
T
1000 MW
200 ns
If the radiation is completely absorbed
without other effects, what is the maximum
height the mass reaches?
Correct answer: 226.757 µm (tolerance ± 1
%).
Explanation:
Let : P = 1000 MW = 1 × 109 W ,
∆t = 200 ns = 2 × 10−7 s ,
m = 10 mg = 1 × 10−5 kg ,
L = 3 cm = 3 × 106 m ,
c = 3 × 108 m/s , and
g = 9.8 m/s2 .
Applying conservation of energy, we obtain
Kf − Ki + Uf − Ui = 0 .
Since Ui = Kf = 0 and Ki =
k~τ k = 1.26454 × 10−12 N · m2 .
The “−” sign indicating the torque points
down, which means the system rotates clockwise, and the direction of the torque is −k̂
(vertically downward).
3
−
−
p2i
, we have
2m
p2i
+ Uf = 0
2m
p2i
+ mgh = 0.
2m
(1)
Using conservation of momentum,
pi = pem =
P ∆t
U
=
.
c
c
(2)
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 4 2007, 11:00 pm
Substituting pi from Eq. 2 into Eq. 1, we
obtain
2
P ∆t
c
−
+ mgh = 0
2m
P 2 (∆t)2
+ mgh = 0.
(3)
2 m c2
Solving Eq. 3 for the maximum height of the
pellet’s trajectory gives us
−
P 2 (∆t)2 1
(4)
2 m2 g c2
(1 × 109 W)2 (2 × 10−7 s)2
=
2 (1 × 10−5 kg)2 (9.8 m/s2 )
1
×
(3 × 108 m/s)2
= (0.000226757 m) (1 × 106 µm/m)
h=
= 226.757 µm .
Question 5, chap -1, sect -1.
part 1 of 1
10 points
A plane electromagnetic wave is generated
due to the initiation of current along the z
direction in a current sheet in the yz plane
at x = 0. A steady flow current is switched
on at t = 0. This current sheet generates a
traveling “front” of the electric and magnetic
fields.
y
infinite
current
sheet in
y, z plane
Px
x
z
Current is in
+z direction, as
shown by arrows.
Consider a new situation, where the current
sheet has the sinusoidal oscillation with an
angular frequency ω. The fields at any point
x and time t are given by:
E = Emax cos(k x − ω t)
B = Bmax cos(k x − ω t)
4
Choose the selection for all possible expressions for the intensity I of the EM waves at
Px .
1
2
c Bmax
only
2 µ0
1
1
2
2
ǫ0 Emax +
c Bmax only
2. I =
2
2 µ0
1
2
c Bmax
only
3. I =
µ0
1
1
2
2
4. I = ǫ0 c Emax
c Bmax
,
,
2
2
µ
0
1
1
2
2
and
ǫ0 Emax +
c Bmax
4
4 µ0
correct
1. I =
2
5. I = ǫ0 c Emax
only
1
1
2
2
6. I =
c Bmax only
ǫ0 Emax +
4
4 µ0
1
2
only
7. I = ǫ0 c Emax
2
1
2
2
8. I = ǫ0 c Emax
,
c Bmax
,
µ
0
1
1
2
2
and
c Bmax
ǫ0 Emax +
2
2 µ0
Explanation:
The time-average of the energy density u at
any point is
1
1
ǫ0 E 2 +
B2 ,
2
2 µ0
= ǫ0 E 2 , or
1 2
B , and
=
µ0
S
I
≡ = .
c
c
u=
For the traveling wave form given, u is independent of x .
I = u c , and
u = uE + uB = 2 uE = 2 uB , so
1
uE = ǫ0 [ Emax cos(k x − ω t)]2
2
1
2
= ǫ0 Emax
.
4
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 4 2007, 11:00 pm
Similarly, uB =
we have
5
1
2
Bmax
. Since uE = uB ,
4 µ0
air
4.
glass
I = 2 uE c =
= 2 uB c =
1
2
,
ǫ0 c Emax
2
1
2
,
c Bmax
2 µ0
Explanation:
When a light ray goes from glass into air, it
refracts away from the normal.
and
= (uE + uB ) c
=
1
1
2
2
ǫ0 c Emax
+
c Bmax
.
4
4 µ0
Question 7, chap 34, sect 3.
part 2 of 2
10 points
air
Question 6, chap 34, sect 3.
part 1 of 2
10 points
glass
Given: A ray approaching an interface.
air
What is the approximate refracted ray?
1. None of these
glass
air
What is the approximate refracted ray?
2.
glass
1. None of these
air
correct
2.
glass
3.
correct
3.
air
glass
air
glass
4.
air
glass
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 4 2007, 11:00 pm
Explanation:
When a light ray goes from air into glass, it
refracts toward the normal.
Question 8, chap 34, sect 3.
part 1 of 2
10 points
Given: The index of refraction of transparent liquid (similar to water but with a
different index of refraction) is 1.22.
A flashlight held under the transparent liquid shines out of the transparent liquid in a
swimming pool. This beam of light exiting
the surface of the transparent liquid makes an
angle of θa = 23 ◦ with respect to the vertical.
flashlight
ray
θa
air
water
θw
6
dent angle in the air is given to be θa = 23 ◦ .
Hence
sin θw
na
=
sin θa
nw
1
sin θw
=
◦
sin(23 )
1.22
0.390731
sin θw =
1.22
θw = arcsin(0.320272)
θw = 18.6793 ◦ .
Question 9, chap 34, sect 3.
part 2 of 2
10 points
The flashlight is slowly turned away from
the vertical direction.
At what angle will the beam no longer be
visible above the surface of the pool?
Correct answer: 55.052 ◦ (tolerance ± 1 %).
Explanation:
This is solved in the same fashion as Part 1.
When the light ceases to be visible outside
the transparent liquid, then θa ≥ 90◦ . The
sin 90◦ = 1. Hence (from above),
na
nw
1
θw = arcsin
1.22
sin θw =
At what angle θw (with respect to the vertical) is the flashlight being held under transparent liquid?
Correct answer: 18.6793 ◦ (tolerance ± 1 %).
Explanation:
Basic Concepts: Snell’s Law:
na sin θa = nw sin θw ,
where na and nw are the indices of refraction
for each substance and θa and θw are the incident angles to the boundary in each medium,
respectively.
Solution: This is a straightforward application of Snell’s law. We assume that the
surface of the transparent liquid is a level horizontal plane, thus each angle with respect to
the vertical represents the incident angle in
each medium.
The index of refraction of air is (nearly)
na = 1.0 while the index of refraction of transparent liquid is given as nw = 1.22. The inci-
θw = 55.052 ◦ .
Question 10, chap 34, sect 3.
part 1 of 1
10 points
A coin is at the bottom of a beaker.
The beaker is filled with 4.8 cm of water (n1 = 1.33) covered by 1.5 cm of
liquid (n2 = 1.4) floating on the water.
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 4 2007, 11:00 pm
How deep does the coin appear to be from
the upper surface of the liquid (near the top
of the beaker)?
Correct answer: 4.68045 cm (tolerance ± 1
%).
Explanation:
Basic Concept: Snell’s Law
n1 sin θ1 = n2 sin θ2 .
θ2 = 0.950001◦
This same ray reaches air at distance
(0.0837758 cm) + (1.5 cm) tan(0.950001◦ )
= 0.108647 cm ,
from the vertical ray, and the angle of refraction is found from
1.4 sin(0.950001◦ ) = 1 sin θ3
For small angles
sin θ ≈ tan θ =
x
.
ℓ
The appearance of the width of the coin x
remains the same. From Snell’s Law, we have
ni
7
x
x
= nf
ℓi
ℓf
Therefore
ℓi
,
ni
since nf ≈ 1 for air and the apparent distance
d = ℓf .
Solution:
ℓf =
l1
l2
+
n2 n1
1.5 cm 4.8 cm
=
+
1.4
1.33
= (1.07143 cm) + (3.60902 cm)
= 4.68045 cm .
d=
The coin appears to be closer to the surface
by
(4.8 cm) + (1.5 cm) − (4.68045 cm)
= 1.61955 cm .
Alternate Solution: Light coming straight
up from the coin falls on each interface at
0◦ and continues straight up. Consider light
making a 1 ◦ angle (therefore we can use the
small angle approximation) with the vertical
in the water. It enters the liquid a distance
θ3 = 1.33◦
Your brain automatically finds the intersection of this ray with the vertical ray, at an
apparent depth of
0.108647 cm
0.108647 cm
≃
◦
tan(1.33 )
0.0232129 rad
= 4.68045 cm .
Rays for all other small angles with the vertical appear to diverge from the image of the
coin at this depth.
Question 11, chap 34, sect 4.
part 1 of 1
10 points
Hint: The convergent mirror in this problem is a part of a lens/mirror system so the
object in this problem may be either real or
virtual. Construct a ray diagram.
Given: A real object is located to the left
of a divergent mirror. The object’s distance
from the mirror and its focal length are shown
in the figure below.
f
◦
(4.8 cm) tan 1 ≃ (4.8 cm) × θ1 u
= 0.0837758 cm
from the vertical ray. In the liquid its angle
with the vertical is given by
◦
1.33 sin 1 = 1.4 sin θ2
13
f
8
0
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 4 2007, 11:00 pm
Which diagram correctly shows the image?
4.
f
1.
f
13
f
8
0
−13
f
21
13
f
8
0
−13
f
21
correct
Explanation:
2.
f
13 −13
f
f0
8
21
p
q f
13
−13
0
f
f
8
21
Basic Concepts:
1 1
1
+ = ,
p q
f
3.
f
13 −13
f
f0
8
21
where f < 0 for a divergent mirror.
Solution:
1
1
1
=− −
q1
p1 f
(8)
1
=−
−
13 f
f
−(13) − (8)
=
13 f
−21
=
13 f
−13
f.
q1 =
21
The magnification m of this mirror is
m=−
q1
p1
8
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 4 2007, 11:00 pm
13
f
= − −21
13
f
8
8
=−
−21
8
=
21
2.
f
13
f
8
Question 12, chap 34, sect 4.
part 1 of 1
10 points
Hint: The convergent mirror in this problem is a part of a lens/mirror system so the
object in this problem may be either real or
virtual. Construct a ray diagram.
Given: A real object is located to the left
of a convergent mirror. The object’s distance
from the mirror and its focal length are shown
in the figure below.
9
13
f
5
0
3.
f
13 13
f
f
5
8
0
correct
f
4.
f
13
f
0
8
Which diagram correctly shows the image?
13
f
8
1.
f
13 13
f
f
5
8
0
Explanation:
0
13
f
5
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 4 2007, 11:00 pm
q
10
h
f
q
p
p
13 13
f
f
0
5
8
Basic Concepts:
1 1
1
+ = ,
p q
f
where 0 < f for a convergent mirror.
Solution:
1 1
1
=− +
q
p f
(8)
1
=−
+
13 f
f
−(8) + (13)
=
13 f
5
=
13 f
13
q=
f.
5
f
Scale: 10 cm =
Find the distance of the image from the
center of the lens.
Correct answer: 17.1107 cm (tolerance ± 1
%).
Explanation:
1 1
1
h′
q
+ =
M=
=−
p q
f
h
p
Convergent Lens
f >0
∞ >p> f
q1
p1
13
f
=− 5
13
f
8
8
=−
5
−8
=
.
5
Question 13, chap 34, sect 5.
part 1 of 2
10 points
A convergent lens has a focal length of
11.7 cm . The object distance is 37 cm .
f <q< ∞
0 > M > −∞
Note: The focal length for a convergent
lens is positive, f = 11.7 cm.
Solution: Substituting these values into the
lens equation
q=
The magnification m of this mirror is
m=−
f h′
=
1
1 1
−
f
p
1
1
1
−
(11.7 cm) (37 cm)
= 17.1107 cm .
Question 14, chap 34, sect 5.
part 2 of 2
10 points
Find the magnification.
Correct answer: −0.462451
%).
Explanation:
M =−
q
p
(tolerance ± 1
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 4 2007, 11:00 pm
=−
(17.1107 cm)
(37 cm)
11
Question 16, chap 34, sect 5.
part 2 of 2
10 points
= −0.462451 .
Find the magnification.
Correct answer: 0.56338 (tolerance ± 1 %).
Question 15, chap 34, sect 5.
part 1 of 2
10 points
A divergent lens has a focal length of 16 cm .
The object distance is 12.4 cm .
Explanation:
q
p
(−6.98592 cm)
=−
(12.4 cm)
M =−
p
q
= 0.56338 .
f
Question 17, chap 35, sect 3.
part 1 of 1
10 points
Scale: 10 cm =
Find the distance of the image from the
center of the lens.
Correct answer: 6.98592 cm (tolerance ± 1
%).
The second-order bright fringe (m = 2) is
4.6 cm from the center line.
1 1
1
h′
q
+ =
M=
=−
p q
f
h
p
Divergent Lens
f <0
∞ >p> 0
S1
θ
S2
1.25 m
f <q< 0 0 <M < 1
Note: The focal length for a divergent lens
is negative, f = −16 cm.
Solution: Substituting these values into the
lens equation
q=
0.0286 mm
Explanation:
1
1 1
−
f
p
1
1
1
−
(−16 cm) (12.4 cm)
= −6.98592 cm
Determine the wavelength of the light. Be
sure to use the small angle approximation,
sin(θ ) ≈ θ
Correct answer: 526.24 nm (tolerance ± 1
%).
Explanation:
=
|q| = 6.98592 cm .
4.6 cm
f
h′
viewing
screen
h
Let : y = 4.6 cm ,
L = 1.25 m , and
d = 0.0286 mm ,
P
−1
an
r2 θ = t
S1
d
y
L
λ
2a
O
Q
S2
δ=
y
r1
viewing
screen
S2 Q S1 ≈ 90◦
6
0
δ ≈ d sin θ ≈ r2 − r1
L
12
+1 λ/d +2 λ/d +3 λ/d
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 4 2007, 11:00 pm
S1
−1
θ = t an
d
θ
S2
y
L
r2
Q
◦
0
≈9
1
r1
S
Q
r2 −
6 S2
≈
nθ
d si
≈
δ
δ=−
How many slits are illuminated?
1. more than eight slits
For constructive interference
ybright =
λL
m,
d
λ
2a
−3 λ/d −2 λ/d −1 λ/d
r1
(1)
with m = 2, y2 = 0.046 m, L = 1.25 m, and
d = 2.86 × 10−5 m
d y2
mL
(2.86 × 10−5 m) (0.046 m)
=
(2) (1.25 m)
= 5.2624 × 10−7 m
= 526.24 nm .
λ=
2. eight slits
3. three slits
4. two slits
5. six slits
6. seven slits
7. five slits correct
8. four slits
Question 18, chap 35, sect 4.
part 1 of 1
10 points
In a diffraction/interference experiment,
identical slits are illuminated by a laser beam.
The intensity pattern shown in the figure is
observed on a screen a large distance from the
slits.
The figure below, shows a plot of the intensity of the light at a distance along the
screen.
9. one slit
Explanation:
Basic Concepts: Multiple Slit Interference and Single Slit Diffraction
This interference pattern (shown in the
question) is the combination of a multiple
five slit interference pattern
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 4 2007, 11:00 pm
and each slit’s
13
If there were three secondary maxima, five
slits are illuminated, etc.
Consequently, in this problem (as shown
in the question) there are three secondary
minima which indicates that there are five
slits illuminated.
Question 19, chap 35, sect 1.
part 1 of 2
10 points
single slit diffraction pattern.
Solution: If only one slit was illuminated,
the intensity pattern is the envelope of the
peak maxima as shown in the figure above,
which is single slit “Fraunhofer” diffraction.
If two slits were illuminated, the intensity
pattern is the envelope of the peak maxima as
shown in the figure below, which is double slit
“Fresnel” interference.
A thin film of cryolite ( nc = 1.36 ) is applied to a camera lens ( ng = 1.53 ). The
coating is designed to reflect wavelengths at
the blue end of the spectrum and transmit
wavelengths in the near infrared.
What minimum thickness gives high reflectivity at 526 nm?
Correct answer: 193.382 nm (tolerance ± 1
%).
Explanation:
For camera lens coating of cryolite (nc =
1.36) over glass (ng = 1.53), high reflectivity
is achieved for
2 nc t1 = m λ1 .
Secondary maxima occur when more than
two slits are illuminated.
If there was one secondary maximum between the central peak and the first large
peak, three slits are illuminated.
If there were two secondary maxima, four
slits are illuminated.
Here we have taken into account that high
reflectivity is achieved for constructive interference. The phase changes at both the “aircryolite” and the “cryolite-glass” surfaces is
φ = 180◦ (nair = 1 < nc < ng ).
Note: Two phase changes of 180◦ .
For minimum thickness m = 1
λ1
2 nc
526 nm
=
2 × 1.36
= 193.382 nm .
t1 =
Question 20, chap 35, sect 1.
part 2 of 2
10 points
What minimum thickness gives high transmission at 1052 nm?
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 4 2007, 11:00 pm
Correct answer: 193.382 nm (tolerance ± 1
%).
Explanation:
Under the same conditions low reflectivity
is achieved for
1
2 nc t = m +
λ2 .
2
For minimum thickness m = 0
λ2
t2 =
4 nc
1052 nm
=
4 × 1.36
= 193.382 nm .
Solution: If the difference in path length
is δ, the phase difference due to the unequal
path lengths is
φpath = k δ =
Interference fringes are produced at point
P on a screen as a result of direct rays from a
605 nm wavelength source and reflected rays
off the Lloyd’s mirror as in the figure. Fringes
0.698 mm apart are formed on a screen 3 m
from the source S.
S1 is the source and S2 is its image.
S2 Q S1 ≈ 90◦
d = 2h
y
r1
−1
an
S1
L
r2 θ = t
d
Q R
δ ≈ d sin θ ≈ r2 − r1
S2
L
sin θ =
θ≈
Explanation:
Basic Concepts: Intensity maxima occur
when the two waves have a phase difference
φ = 0, 2 π, 4 π, · · · = φpath + φref lection ,
y
viewing
screen
O
δ
y
= .
2h
L
(2)
y
δ
≈
2h
L
h=
O
Find the vertical distance h of the source
above the reflecting surface.
Correct answer: 1.30014 mm (tolerance ± 1
%).
Let : L = 3 m
d
h=
2
∆y = 0.698 mm .
P
In the small angle approximation
h
where φref lection = π.
(1)
This diagram shows the similarity between
the Lloyd’s mirror apparatus and Young’s
double slit experiment. The difference between the two experiments is that there is a
180◦ phase change at the reflecting mirror.
viewing
screen
P
Mirror
3m
2π
δ.
λ
6
Question 21, chap -1, sect -1.
part 1 of 1
10 points
S
14
δ
L,
2y
(3)
where the height of the light source above the
d
mirror is h = .
2
So maxima occur at
2π
δ = φ − φref lection
λ
= (0, 2 π, 4 π, · · ·) − π
= −π (below O)
π (1st max)
φpath =
3 π (2nd max), · · · .
The central minimum (at the bottom of the
mirror) is a dark fringe, since the reflected
light produces a 180◦ phase change.
Solution: The separation between maximum
(or minima) on the screen is given in the
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 4 2007, 11:00 pm
problem ∆y = 0.698 mm, and the separation
in path difference is
δ=
λ
2π = λ.
2π
Solving Eq. (3) for h, we have
mδ
L
2 ∆y
mλ
=
L
2 ∆y
1 6.05 × 10−7 m
(3 m)
=
2 (0.000698 m)
= 1.30014 mm ,
h=
where m = 1.
15