1. Shown in the Figure below is a loop of wire that lies in a horizontal plane.
You have a standard bar magnet that you will move either up or down and in the process either insert or remove it from the loop. This problem concentrates on knowing which way the current flows in the loop ( i.e.
all answers are either “clockwise” or “counter-clockwise”). Which way does
N
W E
S the current flow if you:
(a) Insert the North end of the magnet into the loop from below.
ANSWER: clockwise. Inserting the North end of the bar magnet from below increases the magnetic field flux through the loop out of the page. To compensate for this increase, the magnetic field produced by the induced current in the loop has to point into the page. Using your right hand: point the thumb into the page ( this is the direction of B induced
) ). Your curved fingers show the direction of the current that will produce such a field.
(b) Insert the South end of the magnet into the loop from below.
ANSWER: counter-clockwise. Inserting the South end of the bar magnet from below increases the magnetic field flux through the loop into of the page. To compensate for this increase, the magnetic field produced by the induced current in the loop has to point out of the page. Using your right hand: point the thumb out of the page ( this is the direction of B induced
) ). Your curved fingers show the direction of the current that will produce such a field.
(c) Insert the North end of the magnet into the loop from above.
ANSWER: counter-clockwise. Inserting the North end of the bar magnet from above increases the magnetic field flux through the loop into of the page. To compensate for this increase, the magnetic field produced by the induced current in the loop has to point out of the page. Using your right hand: point the thumb out of the page ( this
1

is the direction of B induced
) ). Your curved fingers show the direction of the current that will produce such a field.
(d) Insert the South end of the magnet into the loop from above.
ANSWER: clockwise. Inserting the South end of the bar magnet from above increases the magnetic field flux through the loop out of the page. To compensate for this increase, the magnetic field produced by the induced current in the loop has to point into the page. Using your right hand: point the thumb into the page ( this is the direction of B induced
) ). Your curved fingers show the direction of the current that will produce such a field.
(e) Remove the North end of the magnet from the loop from below.
ANSWER: counter-clockwise. The North pole has been in the loop pointing its flux out of the page. When we remove it from below, we’ll reduce the magnetic flux through the loop pointing out of the page. To compensate for this reduction the magnetic field produced by the induced current needs to point out of the page. Using your right hand: point the thumb out of the page ( this is the direction of B induced
) ). Your curved fingers show the direction of the current that will produce such a field.
(f) Remove the South end of the magnet from the loop from below.
ANSWER: clockwise. The South pole has been in the loop, the
North pole is below the page. When we remove the magnet from below, we’ll reduce the magnet-ix flux through the loop pointing into the page. To compensate for this reduction the magnetic field produced by the induced current needs to point into the page. Using your right hand: point the thumb into the page ( this is the direction of B induced
) ). Your curved fingers show the direction of the current that will produce such a field.
(g) Remove the North end of the magnet from the loop from above.
ANSWER: clockwise. The North pole has been in the loop pointing its flux into the page. When we remove it from above, we’ll reduce the magnet-ix flux through the loop pointing into the page. To compensate for this reduction the magnetic field produced by the induced current needs to point into the page. Using your right hand: point the thumb into the page ( this is the direction of B induced
) ).Your
curved fingers show the direction of the current that will produce such a field.
(h) Remove the South end of the magnet from the loop from above.
ANSWER: counter-clockwise. The South pole has been in the loop pointing the flux out of the page. When we remove it from above, we’ll reduce the magnet-ix flux through the loop pointing out of the page. To compensate for this reduction the magnetic field produced by the induced current needs to point out of the page. Using your right hand: point the thumb out of the page ( this is the direction
2

of B induced
) ).Your curved fingers show the direction of the current that will produce such a field.
2. Shown in the Figure below is a rectangular loop of dimensions 0.20 m by
0.30 m. The loop is partly filled with a magnetic field whose strength is
0.25 T and is directed into the page. The loop has a resistance of 3 Ω.
You grab the loop and pull it to the right with a constant speed of 0.4
m/sec.
x x x x
B=0.25 T x x
B=0 x x x 0.30 m x x x x x x x x x x x x
0.20 m
(a) Which way does the current flow (clockwise or counter-clockwise)?
ANSWER: The loop contains magnetic field pointed into the page.
When you slide it to the right, the downward flux would be reduced.
The loop responds by adding additional downward field lines by making current in the clockwise direction.
(b) What is the voltage generated in the loop?
ANSWER: Here I slipped a jargon ... “voltage” = “emf”. Using
Faraday’s law, we can find the emf. The sign will show us the direction of the current, but we answered that already. So we just need to find the magnitude.
Φ
B
ε = − d Φ
B dt
| ε | = d Φ dt
B
= BA = Blw d Φ
B dt dt d Φ
B
= d ( Blw dt
)
= Bl dw dt
= Blv
= 0 .
25 · 0 .
20 · 0 .
4 = 0 .
02 V olts
(1)
(2)
(3)
(4)
(5)
3

| ε | = 0 .
02 V olts (6)
(7)
(c) What is the current in the loop?
ANSWER:
ε = IR
I =
ε
R
I =
0 .
02
3
= 0 .
0066 Amps = 6 .
6 mA
(8)
(9)
(10)
(d) How much force is required to keep the loop moving at 0.4 m/s?
Let’s first figure out if the magnetic field is exerting force on the loop. The loop has four wire segments, each of which carry the same current. The force on a wire segment with current I is:
−
= I
−
× (11)
One side of the loop is out of the field, so it experiences no force.
The other two sides are parallel to each other and carry current in opposite directions. The forces on these segment will be in opposite directions and will cancel out. For the segment on the left, the force is pointing to the left and will not cancel out. The magnetic field is perpendicular to the wire, so the cross product gives:
F = ILB = 0 .
0066 · 0 .
20 · 0 .
25 = 0 .
000333 N ewtons (12)
4

3. Shown in the figure below is the view from above of a sliding bar of resistance R and length L. The bar is free to slide along the rails without friction. The system is immersed in a magnetic field pointed out of the page (toward you). The bar is being pulled with constant velocity v to the bottom of the page. Answer all the following:
(a) Write an expression for the flux through the loop formed by the bar and wire slide for the moment when the bar’s position is “y”.
ANSWER:
Φ
B
= BA = BLy (13)
(b) Using your result from (a) determine an expression for the voltage generated in the loop as a function of the bar’s velocity.
ANSWER: The velocity
V = − d Φ
B dt
V = − Bl dy dt
= − BLv
(14)
(15)
(c) Determine the current through the bar.
ANSWER:
V = IR
I =
V
R
I = −
BLv
R
(16)
(17)
(18)
5

The magnetic flux is increasing out of the page, so the induced current needs to flow clockwise to create magnetic field flux into the page.
(d) Determine the force on the bar as a function of velocity.
ANSWER: The field is perpendicular to the bar, so the magnitude of the force is:
F = ILB
F =
F =
BLv
R
LB
R
B 2 L 2 v
(19)
(20)
(21)
(e) What direction is this force (justify your response).
ANSWER: The force is in the plane of the page pointing up (opposing the change). This can be seen from the vector product: current to the left, field out of the page, force - up.
4. The circuit below begins with the switch in the open position. The battery has V
B
= 8 V , the resistor has R = 2 k Ω, and the inductor has L = 3 mH .
At time=0, the switch in the circuit is closed.
(a) Determine the time constant of this circuit.
ANSWER:
τ =
L
R
=
3 mH
2 k Ω
= 1 .
5 µsec
(b) Draw sketches of the time dependence of each of the following: i. The voltage on the inductor as a function of time, V
L
( t ).
(22)
8 V
V L t
6

8 V
V R t
I
R
4 mA t ii. The voltage on the resistor as a function of time, V
R
( t ).
iii. The current through the resistor as a function of time, I
R
( t ).
To be counted for full credit the vertical axis of each sketch should be labelled with a numerical value indicating either the initial or asymptotic value of the quantity plotted.
(c) Write an equation for each of the following:
ANSWER: i. The voltage on the inductor as a function of time, V
L
( t ).
V
L
( t ) = 8 V e − t
1 .
5 µsec
(23) ii. The voltage on the resistor as a function of time, V
R
( t ).
V
R
( t ) = 8 V 1 − e − t
1 .
5 µsec
(24) iii. The current through the resistor as a function of time, I
R
( t ).
I
R
( t ) = 4 mA 1 − e − t
1 .
5 µsec
(25)
(d) At what time is the voltage on the inductor 3 V?
3 V = 8 V e − t
1 .
5 µsec
0 .
375 = e − t
1 .
5 µsec ln 0 .
375 = − t
1 .
5 µsec t = − 1 .
5 µsec × ln 0 .
375 = 1 .
47 µsec
(26)
(27)
(28)
(29)
7

5. A radio tuning circuit consists of a variable capacitor and an inductor.
The minimum value of the capacitor is C = 4 .
18 pF .
(a) If the frequency on one end of the AM radio broadcast band is f = 1 .
6 M Hz and you want to make a circuit that can receive it by tuning the capacitor at its minimum value, find the inductance of the coil that you will need.
ANSWER: Here is an easy one. The resonance frequency of the LC circuit depends on L and C.
ω = 1 /
√
LC (30)
LC = 1 /ω 2 (31)
L = 1 / ( C · ω 2 ) (32)
Now, ω is the angular frequency and we are given the frequency f .
How do I know that ? Because of the units - Hz means 1 period per second. The angular frequency is measured in rad/s , i.e. - angle per unit time. Since 1 period is equal to an angle of 2 π rad , ω = 2 π f .
L = 1 / (4 .
18 × 10 −
12
L = 1 / ( C · 4 · π
2
· 4 · π 2 · (1 .
6 × 10
·
6 f
2
) 2
)
)
L = 2 .
37 mH
(33)
(34)
(35)
(b) If the other end of the AM radio broadcast band is f = 540 kHz what is the maximum value of the adjustable capacitance ?
ANSWER: This is also easy. The same relations - we just change the value of the frequency, so we need to change something in the circuit.We’ll change the capacitance. You can imagine that by turning the knob on your radio, you are changing the overlap area of the plates of the capacitor. An then ...hey you are tuning the resonant frequency! So, here is what we need for C to get to the other end of the AM range:
C = 1 / [ L (2 π f ) 2 ]
C = 1 / [2 .
37 × 10
−
3 (2 π · 540 × 10 3 ) 2 ]
C = 36 .
7 pF
(36)
(37)
(38)
I can do it in another way, not involving a numeric value for L at all.
C new
= 1 / [ L (2 π f new
) 2 ]
C new
= C old
(2 π f old
) 2 / (2 π f new
) 2
C new
= C old
· f 2 old f 2 new
(39)
(40)
(41)
(c) With the capacitor tuned as in part (a) the maximum current through the circuit is I = 0 .
85 mA . What is the maximum charge on the capacitor ?
8

ANSWER: Here we will use conservation of energy. The energy will be converted from magnetic energy stored in the inductor to electric energy stored in the capacitor and vice versa. We know the maximum current,L and C. So we know the total energy of the system.
U =
1
2
LI 2 max
Q max
=
1
2
Q 2 max
= I max
/C
√
LC
(42)
(43)
Here, I can type in the numbers for L and C... but I’m lazy and I notice that:
√
LC = 1 /ω = 1 / (2 π f ) (44)
So, now I need to type less numbers, because I’m given f = 1 .
6 M Hz :
Q max
= I max
/ (2 π f )
Q max
= 0 .
85 × 10
−
3 / (2 π 1 .
6 × 10 6 )
Q max
= 84 .
5 pC
(45)
(46)
(47)
(d) Find the wavelengths that correspond to the two ends of the radio broadcast band.
ANSWER: OK, whew! A very easy one: c = λf
λ = c f
3 × 10 8
λ
1
=
λ
2
=
540 × 10 3
3 × 10 8
1 .
6 × 10 6
= 555 .
5 m
= 187 .
5 m
(48)
(49)
(50)
(51)
(52)
9

6. You were outside your family spacecraft taking a leasurely spacewalk and your impetulant kids speed off into the cosmos leaving you just floating there, about 150 billion meters from the sun. At your location, the intensity of sunlight is roughly 1000 W m 2 and the gravity is not completely zero, so you start to fall toward the sun. Even though it would take a long time to reach the sun, you are concerned that if your position changes alot you’ll never be found.
So you take action:
(a) You calculate the average electric field of the light.
ANSWER: OK, now you should remember that the intensity ( W m 2
) is related to the magnitude of the Pointing vector.
Thus:
I = ǫ
0
E 2 × c r I
E = ǫ
0 c
E = r 1000
8 .
85 × 10
−
12 · 3 × 10 8
E = 614
V m
(53)
(54)
(55)
(56)
(b) You calculate the average magnetic field of the light.
ANSWER: I can write this in terms of the magnetic field:
B 2
I =
µ
0
× c
B = r
µ
0 c
I
B = r
4 π × 10
−
7 ·
3 × 10 8
1000
B = 0 .
00000205 T esla
(57)
(58)
(59)
(60)
Hey, I can do this much easier. Aren’t the magnetic field and electric field in an EM wave proportional via:
E = cB
B =
E c
B =
614
3 × 10 8
= 0 .
00000205 T esla
(61)
(62)
(63)
Um, yup :)
10

(c) You calculate the pressure applied to a reflective solar sail.
ANSWER: The pressure applied by EM waves varies from P = I c to P = 2 I c as the substance to which the light is applied goes from absorbative to reflective. Yours is reflective:
P =
2
2000 c
I
P = 0 .
P =
00000666
3 × 10 8
P a (
N m 2
)
(64)
(65)
(66)
(d) You calculate the size of a solar sail necessary to exactly balance the sun’s gravity.
ANSWER: Well, due to gravity you get F = G mM r 2
. Due to the solar wind, you get F = P ressure × Area . Set these equal to avoid falling into the sun!
A =
6 .
67 × 10
−
11 ·
G mM r 2
A
100 ·
=
2
=
×
P A
GmM
P r 2
10
0 .
00000666 · (150 × 10 9 ) 2
30
A = 88942 m 2
(67)
(68)
(69)
(70)
As a square, this would be 300 meters on each side. See how bad the Star Wars Part-II Solar Sail was...WAY WAY too small.
(e) What do you write on the sail?
ANSWER: Your situation is rather hopeless, so I leave it up to you to figure out the best strategy.
Oh, you and your space suit have a total mass of 100 kg and you happen to remember that the mass of the sun is 2 × 10 30 constant G = 6 .
67 × 10
−
11 N m 2 /kg 2 .
kg and the gravitational
11

7. Shown in the figure below is a system containing an object, and two lenses.
Use the shapes of the lenses in the figure to decide whether they are converging or diverging optical elements.
|f|=8 cm |f|=30 cm
15 cm 4 cm
(a) Find the image location and magnification of the first lens (assuming that only this lens exists). Specify this image location, d i 1
, as some number of centimeters to the left or to the right of this lens.
ANSWER: d o 1
= 15 cm f
1
= 8 cm
1
15 d i 1
+ d
1 i 1
=
1
8
= 17 .
14 cm m
1
= −
17 .
14 cm
15 cm
= − 1 .
143
(71)
(72)
(73)
(74)
(right of lens)
(b) The image of the first lens acts as the object for the second. Find the location and magnification of the image produced by the second lens. Specify this image location, d i 2
, as some number of centimeters to the left or to the right of this lens.
ANSWER: d o 2
= − 13 .
14 cm
1
− 13 .
14
+ d i 1 f
2
= − 30 cm d
1 i 2
=
1
− 30
= 23 .
38 cm m
1
= −
23 .
38 cm
− 13 .
14 cm
= 1 .
779
(right of lens)
(75)
(76)
(77)
(78)
12

(c) Calculate the total magnification of this entire system.
ANSWER: m
T OT
= m
1
× m
2
= − 2 .
034
(d) Is the final image real or virtual?
ANSWER: real
(79)
8. Shown in the figure below is a system containing an object, a lens, and a mirror. This is a 3-pass system in that light goes from the object, through the lens, off the mirror, and back through the lens a second time .
Use the shape of the objects in the figure to decide whether they are converging or diverging optical elements.
|f|=5 cm R = 10 cm
20 cm 8 cm
(a) Find the image location and magnification of the first lens (assuming that only this lens exists).
ANSWER: Because the object is on the incoming side, d o 1 is positive, i.e.
d o 1
= +20 cm . Because the lens is diverging, the focal length is negative, i.e.
f
1
= − 5 cm . Using 1 d o 1
+ 1 d i 1
= 1 f
1
, we have m
1
= − d i 1 d o 1
1
20
1 d i 1
+
= d
1 i 1
1
− 5
=
−
1
− 5
1
20
1 d i 1
= − 0 .
25 d i 1
= −
− 4
20
= − 4 cm
= +0 .
2
(80)
(81)
(82)
(83)
(84)
13

The negative sign for d i 1 means that the image is NOT on the outgoing side, so the image is 4 cm to the left of the lens on the first pass.
(b) The image of the lens acts as the object for the mirror. Find the location and magnification of the image produced by the mirror.
ANSWER: Since the image of the lens is 4 cm to the left of the lens itself, the object distance for the mirror is 8 cm + 4 cm = 12 cm . This distance is positive since this “object” is on the incoming side of the light, i.e.
d o 2
= +12 cm . The focal length of a curved mirror is 1
2 of the radius. Also, since this mirror is a converging device, the focal length is positive. Thus, f
2
= +5 cm . Plugging these in: m
2
= − d i 2 d o 2
1
12
1 d i 2
+
=
1 d i 2
1
5
−
=
1
1
5
12 d
1 i 2
= 0 .
11666 d i 2
= 8 .
5714 cm
= −
8 .
5714
12
= − 0 .
714
(85)
(86)
(87)
(88)
(89)
Since this image distance is positive, the image is on the outgoing side of the mirror at a distance of 8.57 cm. The outgoing side of the mirror is to the left of the mirror.
(c) Lastly, the image from the mirror acts as the object for the light’s second pass through the lens. Find the location and magnification of the image produced in this step.
ANSWER: This one is a bit tricky. Since the image of the mirror
(object for the lens) is 8.5714 cm away from the mirror and the lens is only 8 cm from the mirror, the object for the lens is NOT on the incoming side. Thus, the object distance will be negative, d o 3
= − 0 .
5714 cm . Again, the lens is diverging, so that the focal length is f = − 5 cm . OK, putting this in: m
3
= − d i 3 d o 3
1
− 0 .
5714 d
1 i 3
=
1
− 5
+
−
1 d i 3
− 0
=
.
1
1
− 5
5714 d
1 i 3
= 1 .
55 d i 3
= 0 .
645 cm
= −
0 .
645
− 0 .
5714
= 1 .
129
14
(90)
(91)
(92)
(93)
(94)

This image is on the outgoing side meaning to the left of the lens.
(d) Calculate the total magnification of this entire system. Is the final image real or virtual?
ANSWER: m
T OT
= m
1
× m
2
× m
3
= 0 .
2 ×− 0 .
714 × 1 .
129 = − 0 .
161
This image is real since it is on the outgoing side of the last optical element, and the minus sign tells us that it is inverted as it should be for real images.
9. A particular person’s nearsighted eye has a near point of 12 cm and a far point of 17 cm. You will place a lens 2 cm in front of their eye. Answer all the following:
(a) Do you use a converging or diverging lens?
ANSWER: This near-sighted person should have a diverging lens.
(b) What is the focal length of the chosen lens?
ANSWER: Well, this person has a far point of 17 cm. Normally, the eye should have a far point of infinity. So, what we do with the lens is we take objects at infinity, and put the image 17 cm in front of this person’s eye.
Now, since the lens is not touching the eye (these are glasses not contacts), we need an image that is 17-2=15 cm from the lens. Note that this image will NOT be on the outgoing side, so we conclude d i
= − 15 cm when d o
= ∞ .
1
∞ d
1 o
+
+
1
1 d i
− 15
1
−
15 f = − 15 cm
=
1
= f
1
= f
1 f
(95)
(96)
(97)
(98)
(c) What is this person’s new nearpoint?
ANSWER: Well, this person will reach the nearpoint when the image from the glasses is 12 cm in front of the eye, or 10 cm in front of the lens d i
= − 10 cm .
1 d o
1
+ d o
+
1
− 10
1 d i
=
=
1
1 f
− 15
(99)
(100)
15

1 d o
= d o
1
10
−
1
15
= 30 cm
(101)
(102)
Wait, what if this poor person wants to see an object when it is closer than 30 cm? Well, they take off their glasses!! How many times have you seen a person look at something closely by taking off their glasses? Each such time, you were witnessing a near-sighted person.
10. Shown in the figure below is a water glass at two different moments. In the first moment (left figure) the water glass is empty and the light ray strikes the bottom corner as shown. In the second moment, the glass has been filled with water (n=1.33). As a result, the light ray has been bent.
Light Ray
θ
1
Light Ray
θ
1 n=1.33
θ
2
15 cm
Empty water glass
8 cm
Full water glass x
(a) Determine the angle θ
1 as shown in the left figure.
ANSWER: Examining the figure closely, we see that the angle θ
1 is also the upper angle in the triangle formed by the right ray and the left and bottom sides of the glass (using vertical angles). In this case, we see that tan θ
1
= 8 cm
15 cm
. Solving this we find that θ
1
= 28 .
1 o .
(b) Determine the angle θ
2 as shown in the right figure.
ANSWER: Using Snell’s law: n
1 sin θ
1
= n
2 sin θ
2
1 × sin 28 .
1 o
= 1 .
33 × sin θ
2
0 .
354 = sin θ
2
θ
2
= 20 .
74 o
(103)
(104)
(105)
(106)
(c) Determine the distance x shown in the right figure.
16

ANSWER: Examining the triangle, we see the following: tan θ
2
= x
15 cm x = 15 cm × tan θ
2 x = 15 cm × tan 20 .
74 o x = 5 .
68 cm
(107)
(108)
(109)
(110)
11. As shown in the figure below, a small boy has lost his toy at the bottom of a pool ( n water
= 1 .
33). When he shines his flashlight as shown, he can see the toy.
1 m
1.5 m
3.0 m
θ
1
θ
2 x
(a) What is the angle of incidence, θ
1
, of his light upon the water?
ANSWER: tan
θ
1
θ
1
=
3 .
0
1 .
0
= 71 .
6 o
(111)
(112)
(b) What is the angle of refraction θ
2
?
ANSWER: n
1 sin θ
1
= n
2 sin θ
2
1 .
0 sin 71 .
6 o
= 1 .
33 sin θ
2
θ
2
= 45 .
5 o
(c) What is the distance from the water’s edge, x , to the lost toy?
ANSWER:
(113)
(114)
(115) x = 3 m + l tan θ
2
= l
1 .
5 l = 1 .
5 tan θ
2 l = 1 .
5 tan 45 .
5 o l = 1 .
53 m x = 4 .
53 m
(116)
(117)
(118)
(119)
(120)
(121)
17

12. Shown in the figure below is Young’s two-slit experiment. Monochromatic light of wavelength λ = 0 .
633 µm comes in from the left and passes through the two (very narrow) slits. The result is a pattern of bright and dark bands visible on a screen 5 meters away from the two slits.
WARNING: this drawing is not to scale.
d
3 cm
Dark
Bright
Dark
Bright
5 meters
(a) Determine the distance, d, between the two slits.
ANSWER: The distance labelled in the figure (3 cm) is the distance to the first bright band. This band follows the relation d sin θ = λ . The angle to this band can be found as follows: tan θ =
3 cm
5 m
=
0 .
03
5 m m
θ = 0 .
344 o d =
λ sin θ d sin θ = λ
=
0 .
633 µm sin 0 .
344 o
= 105 .
5 µm
(122)
(123)
(124)
(125)
(b) If the light is changed to green light, λ = 0 .
532 µm , the interference pattern will change. What is the location, y, of the first dark band using this light.
ANSWER: The dark bands follow the relation d sin θ = n + where n = 0 , ± 1 , ± 2 , ...
. The first dark band is n = 0 or d sin θ =
1
2
1
2
Using this with the new wavelength yields:
λ
λ .
d sin θ =
1
2
λ
105 .
5 µm sin θ = sin θ =
1
1
2
2
×
×
0
0 .
.
532
532
105 .
5
µm
µm
µm sin θ = 0 .
00252 y = 5 meters × tan 0 .
1445 o sin θ = 0 .
00252
θ = 0 .
1445 o tan θ = y
5 meters
= 0 .
0126 m = 1 .
26 cm
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(131)
(132)
(133)
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(c) If the slits have finite width a = 10 µm then the intensity in the intereference maxima will change and a new structure will appear due to the diffraction. For λ = 0 .
532 µm , find the location of the first dark diffraction band.
ANSWER: Now, because of the finite width, we need to consider the diffraction. Since the slit is much narrower than the slit spacing, we expect that the structure that will appear will have larger size on the screen and it will give the intensity envelope for the interference maxima, i.e. they will no longer have equal intensity, as shown in the figure. The condition for a minimum from a single slit diffraction is: a sin θ = nλ n = 1 sin θ =
0 .
532 µm
10 µm
θ = 3 .
0452 o y = 5 meters × tan 3 .
0452 o
= 0 .
266 m = 26 .
6 cm
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13. A thin film of soap water (n=1.33) appears green when lit from above.
Calculate the thickness of the film if it appears green ( λ = 500 nm ).
Assume that this is the thinnest film that can produce green light.
ANSWER: OKAY. Each time light is reflected from an interface between two materials of different index of refraction, there is a chance there will be a phase change in the light. If the second medium is higher n, the phase change is ∆ φ = π . However, if the second medium is lower n, the phase change is ∆ φ = 0.
So, if you have a film with nearly zero thickness, the reflections from the two surfaces cancel (watch the top of a soap bubble just before it bursts, the reflections go away!). Now, we *want* reflections, so the film should not have zero thickness. Instead the round-trip distance of the light should be 1/2 wavelength so that the destructive turns to constructive.
Finally, the 1/2 wavelength should be 1/2 of the wavelength while the light is in the soap film !!
d =
1
4
2 d =
λ n
=
1 λ
2 n
λ
4 n d = 94 nm
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14. Assume that three perfect polarizers are placed in sequence and receive incoming light from one side. The first polarizer has it’s optic axis oriented vertically. The second is oriented 30 degrees to one side of vertical. The third is oriented horizontally.
(a) What fraction of the incoming light gets through the first polarizer?
For any polarizer the outgoing intensity is related to the incoming intensity via the formula I out
= I in cos 2 φ . Here φ is the angle between the polarization of the light and the axis of the polarizer. Oops!
In the first stage the incoming light is unpolarized. This means φ takes all possible angles. However, we know about the average value: cos 2 φ = 1
2
. Thus:
I
1
= I
0 cos 2 φ
F raction
I
1
=
I
I
= I
0
1
1
0
=
2
1
2
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(144)
(b) What fraction of the incoming light gets through both the first and second polarizer?
In this case the calculation is simpler:
I
2
= I
1 cos 2 30 o
I
2
= I
1 cos 2
= I
1
√
3
2
!
φ
2
F raction =
I
2
I
0
I
2
=
I
2
=
= I
1
3
I
0
1
2
4
3
1
2
3
4
=
4
3
8
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(149)
(c) What fraction of the incoming light gets through the first, second, and third polarizers?
Again, a simple case:
I
3
= I
2 cos 2 60
I
3
= I
2 cos 2 φ
2 o = I
2
1
2
I
3
= I
2
1
4
20
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(152)

I
2
F raction =
I
0
=
I
3
= I
0
3
3
1
8
3
4
8
1
4
=
32
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(154)
(d) How much light would get through if the second and third polarizers were exchanged?
Nothing!!
(e) Which way are the polarizers in your sunglasses oriented. Explain why.
Glare comes mostly from reflections off of horizontal surfaces. This glare is highly polarized parallel to the surface ( i.e.
horizontally).
To cut the glare, you orient the axis of your polarizing singlasses vertically.
(f) Explain how polarizing glasses are used to view high quality 3D movies.
3D movies operate on the principle of sending different images into each eye. The images are selected so that they view the same scene from different angles. To make a high quality 3D movie, you use two psojectors sending light to the same screne. The two projectors send polarizations that are perpendicular to each other. By using “special glasses” ( i.e.
with polarizing filters over the two eyes) you can see the depth of the image. All one does is match the polarization direction of projector 1 to the left eye, and projector 2 to the right eye. Bottom line is different images to different eyes and a 3D movie results.
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