midterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 6 2007, 11:00 pm
E & M - Basic Physical Concepts
Current and resistance
Current: I = ddtQ = n q vd A
Ohm’s law: V = I R, E = ρJ
I , R = ρℓ
E = Vℓ , J = A
A
Electric force and electric field
Electric force between 2 point charges:
|q | |q |
|F | = k 1r2 2
k = 8.987551787 × 109 N m2 /C2
ǫ0 = 4 π1 k = 8.854187817 × 10−12 C2 /N m2
qp = −qe = 1.60217733 (49) × 10−19 C
mp = 1.672623 (10) × 10−27 kg
me = 9.1093897 (54) × 10−31 kg
~
~ =F
Electric field: E
2
Power: P = I V = VR = I 2 R
Thermal coefficient of ρ: α = ρ ∆ρ
0 ∆T
Motion of free electrons in an ideal conductor:
a τ = vd → qmE τ = nJq → ρ = n qm2 τ
|Q|
~2 + · · ·
~ =E
~1 + E
Point charge: |E| = k r2 , E
Field patterns: point charge, dipole, k plates, rod,
spheres, cylinders,. . .
Charge distributions:
Linear charge density: λ = ∆Q
∆x
Surface charge density: σsurf =
∆Q
Volume charge density: ρ = ∆V
∆Qsurf
∆A
Electric flux and Gauss’ law
~ · n̂∆A
Flux: ∆Φ = E ∆A⊥ = E
Gauss law: Outgoing Flux from S, ΦS = Qenclosed
ǫ0
Steps: to obtain electric field
~ pattern and construct S
–Inspect E
H
~ · dA
~ = Qencl , solve for E
~
–Find Φs = surf ace E
ǫ
0
Spherical: Φs = 4 π r2 E
Cylindrical: Φs = 2 π r ℓ E
Pill box: Φs = E ∆A, 1 side; = 2 E ∆A, 2 sides
σ
k
~ = 0, Esurf
Conductor: E
= 0, E ⊥ = surf
in
surf
Potential
ǫ0
Potential energy: ∆U = q ∆V 1 eV ≈ 1.6 × 10−19 J
Positive charge moves from high V to low V
Point charge: V = krQ V = V1 + V2 = . . .
1 q2
Energy of a charge-pair: U = k rq12
Potential difference: |∆V | = |E ∆sk |,
R
~ · ∆~s, V − V = − B E
~ · d~s
∆V = −E
B
A
A
¯
¯
d
V
∆V
E = − dr , Ex = − ∆x ¯
= − ∂V
∂x , etc.
f ix y,z
Capacitances
Q=CV
Series: V = CQ = CQ + CQ + CQ + · · ·, Q = Qi
eq
1
2
3
Parallel: Q = Ceq V = C1 V + C2 V + · · ·, V = Vi
ǫ A
Q
Parallel plate-capacitor: C = V
= EQd = 0d
2
RQ
Q
Energy: U = 0 V dq = 12 C , u = 12 ǫ0 E 2
2
1
2
Uκ = 21κ Q
C0 , uκ = 2 ǫ0 κ Eκ
Q
Q
Spherical capacitor: V = 4 π ǫ r1 − 4 π ǫ r2
0
0
~
Potential energy: U = −~
p·E
Dielectrics: C = κC0 ,
V =IR
Direct current circuits
q
Area charge density: σA = ∆Q
∆A
1
Series: V = I Req = I R1 + I R2 + I R3 + · · ·, I = Ii
V + V + V + · · ·, V = V
Parallel: I = RV = R
i
R2
R3
eq
1
Steps: in application of Kirchhoff’s Rules
–Label currents: i1 , i2 , i3 , . . .
P
P
i
–Node equations:
i =
P in
Pout
–Loop equations: “ (±E) + (∓iR)=0”
–Natural: “+” for loop-arrow entering − terminal
“−” for loop-arrow-parallel to current flow
RC circuit: if ddty + R1C y = 0, y = y0 exp(− RtC )
Charging: E − Vc − R i = 0, 1c ddtq + R ddti = ci + R ddti = 0
Discharge: 0 = Vc − R i = qc + R ddtq , ci + R ddti = 0
Magnetic field and magnetic force
µ0 = 4 π × 10−7 T m/A
µ a2 i
µ i
0
Wire: B = 2 π0 r
Axis of loop: B =
2 (a2 +x2 )3/2
~ → q ~v × B
~
Magnetic force: F~M = i ~ℓ × B
~ × B,
~
Loop-magnet ID: ~τ = i A
µ
~ = i A n̂
2
r
Circular motion: F = mrv = q v B, T = f1 = 2 π
v
~ + q ~v × B
~
Lorentz force: F~ = q E
~
Hall effect: V = FM d , U = −~
µ·B
H
q
~ and magnetism of matter
Sources of B
µ
~
µ
v ×r̂
0 q~
~ = 0 i ∆ℓ×r̂
Biot-Savart Law: ∆B
4 π r2 , B = 4 π r2
2
µ0 i ∆y
∆B = 4 π
sin θ, sin θ = ar , ∆y = r a∆θ
r2
H
~ · d~s = µ I
B
Ampere’s law: M =
L
0
encircled
Steps: to obtain magnetic field
~ pattern and construct loop L
–Inspect B
~
–Find M and Iencl , and solve for B.
d (E A)
ΦE = ǫ
Displ. current: Id = ǫ0 d dt
0
dt
Magnetism in atom:
Orbital motion: µ = i A = 2 em L
L = m v r = n h̄,
QA
= d dt
h̄ = 2hπ = 1.06 × 10−34 J s
h̄ = 9.27 × 10−24 J/T
µB = 2em
µspin = µB
Magnetism in matter:
0
B = B0 + BM = (1 + χ) B0 = (1 + χ) µ0 B
µ0 = κm H
Ferromagnetic: χ ≫ 1
Diamagnetic: −1 ≪ χ < 0
Paramagnetic: 0 < χ ≪ 1, M = C
TB
µorbit = n µB ,
Spin: S = h̄2 ,
midterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 6 2007, 11:00 pm
Complete reflection: P = 2cU ,
Faraday’s law
φB
E = −N ddt
,
φB =
R
~M
~ = F
E
q
~ · dA
~,
B
d (B A⊥ )
d φB
A⊥
= ddtB A⊥ + B d dt
dt =
dt
´
³
d 1 R ·Rθ
Moving rods: ddtA = ℓ v, ddtA = dt
2
A⊥ = d (A cos ωt)
Rotating loop: d dt
dt
Cutting B lines → change φB → Eind → Eind
Maxwell equations:
H
~ · dA
~ = Q,
E
ǫ0
H
φB
~
,
E · d~s = − ddt
H
H
~ · dA
~ = 0,
B
~ · d~s = µ [I + ǫ d φE ]
B
0
0 dt
Inductance
M21 = M12 = N2i φ21
1
L = Ni φ , VL = L ddti
Mutual: E2 = −M21 ddti1 ,
Self: E = −L ddti ,
A
Long solenoid: L = N B
i , B = µ0 n i
Energies: UL = 21 L i2 , uB = 2 1µ B 2
0
UC = 21C q 2 , uE = 21 ǫ0 E 2
q
q = q0 cos(ω t + δ),
L C: VL + VC = 0 ⇒ L ddti = − C
q
ω = L1C , UC + UL = UC max = UL max = U0
Decay Equations: ddty = −a y, y = y0 exp(−a t)
VL + R VL = 0,
L R: E = VL + R i, ddt
Lh
´
³
´i
³
E 1 − exp −R t
,
i
=
VL = E exp − Rt
L
R
L
L R C:
r
³ ´2
R
Q ≈ Q0 e− 2 L t cos ωd t, ωd = L1C − 2RL
Underdamped, critically damped & overdamped
A C Circuits
q
Impedance: [Ohm ≡ Ω]
Z≡
R2 + (XL − XC )2
Inductive XL = ω L, Capactive XC = ω1C
R
Mean value: f¯(t) = T1 0T f (t) dt
1
1
2
2
[sin ω t]rms = [sin2 ω t] = [ 12 (1 − cos 2 ω t)] = √1
2
Electromagnetic waves
Properties of em waves:
E = Em cos(k z − ω t), B = E
c
λ
c
v = ddtz = ω
k = λf = T , n = v
speed of light: c = √ǫ1 µ = 2.99792458 × 108 m/s
0
P = 2cS
Reflection and Refraction
~ · d~s,
E= E
~ opposes change of Φ
Lenz law: Induced B
B
R
2
0
~ ⊥ E,
~ propagating along: E
~ ×B
~
B
u = uE + uB , uE = uB
~ B
~
~ = E×
Poynting vector: S
, S̄ = I¯ = Erms Brms
µ0
µ0
∆U
d
z
P
Intensity: I = A = A ∆z dt = u c
R
~ · dA
~ = dU + P
Energy conservation:
S
R
dt
Complete absorption: Momentum p = Uc
∆p 1
∆U 1
S
Pressure: P = F
A = ∆t A = c ∆t A = u = c
n1 = v2 = λ2
Index of refraction: n
v1
λ1
2
Snell’s law: n1 sin θ1 = n2 sin θ2
Critical angle: n2 > n1 , n2 sin θc = n1 sin 90◦
Total reflection: θ > θc
Mirrors and lenses
1
1
1
p+q = f
Ray tracing rules:
Mirror: At symm pt S, reflected symmetrically through
center of sphere, undeflected. Parallel to axis, converges
toward F (or diverges away from F ), f = R
2 .
Lens: Through center of lens, undeflected. Parallel to
axis, converges toward F (or diverges away from F )
Image: q > 0 (real), q < 0 (virtual)
Focal point F : at p = ∞, q = f
f = ±|f |, “+” convergent, “−” divergent
′
Magnification: M = hh = − pq
1
Refraction at spherical surface: np1 + nq2 = n2 −n
R
R is coordinate of center with origin at S, with
S the symmetry³point of´ ³
surface on ´the axis
n
1
2
Lens maker: f = n1 − 1 R1 − R1
1
2
′
1
Two media: M = hh = − pq n
n2
Huygen’s principles:
Points in wave front are sources of next wavelets
Forward tangent surface is next wave front
Interference
Maxima φ = 0, 2 π, 4 π, · · ·; Minima
³ φ´ = π, 3 π, 5 π, · · ·
Double slits: Iaverage = I0 cos2 φ
2 , φ = k∆.
y
for small θ, θ ≈ sin θ ≈ tan θ
sin θ = ∆
d , tan θ = L ,
~=A
~1 + A
~2 + A
~3 + · · ·
Phasor diagram: A
Ax = A1x +A2x +A3x +· · ·, Ay = A1y +A2y +· · ·
a
b
c
sin α = sin β = sin γ
First minimum for N slits: φ = 2Nπ
Thin film: φ = k ∆ + |φ1ref lected − φ2ref lected |, ∆ = 2 t
φref lected = π (denser medium); =0 (lighter medium)
·Diffraction
¸2
sin β2
Single slit: I = I0
, β = k∆,
β
∆ = a sin θ
2
λ
Resolution criterion: θcriterion = 1.22 D
Grating: Principle maxima ∆ = m λ
Polarization
Brewster (n1 < n2 ): n1 sin θbr = n2 sin( π2 − θbr )
Polarizer: Etransmit = E0 cos θ, I = I0 cos2 θ
I0
Unpolarized light: ∆I
∆θ = 2 π
Transmitted Intensity: ∆I ′ = ∆I cos2 θ
R
I ′ = 2I0π 02 π cos2 θ dθ = I20
midterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 6 2007, 11:00 pm
3
.
Flow Diagram for iCliker test Mode
BEFORE CLASS: log into TT, go to menu item A.4 \manage aount info". Enter the serial number (eight haraters)
from your liker, and memorize your box number.
quiz mode
Is the exam test or quiz mode?
test mode
quiz mode is easy.
Just
bring your liker to lass.
In lass: using the box number (above),
nd your box on the projeted sreen.
re-enter
From the top of page 1 of your test, get the 5-letter version ode. Type this
ode using your liker. DO NOT PRESS THE \E" (enter) BUTTON YET.
no
Version
Entry
Does the ode in your box agree with the version ode printed on your test?
yes
DO NOT PRESS THE
\E" (enter)
BUTTON.
Press \E" to onvert the version ode to a number.
no
Does the version number in your box (1, 2, or 3 digits) agree with
the version number printed at the top of page 1 of your test?
yes
Press \E" to submit your version number.
Use the \CC" (previous question) and \CD" (next question)
odes to navigate to whih question you want to answer. These
two odes must be submitted using the \E" (enter) button.
Find and enter the 2-letter hoie ode for your answer from the \TWO-LETTER CODES" listed below. DO NOT PRESS \E" (enter) BUTTON YET.
Answer
Entry
Does the hoie ode in your box agree with your entry?
yes
Press \E" to onvert the hoie ode to a hoie number.
Does the number in your box agree with your seleted hoie?
yes
Press \E" (enter) button to submit your hoie.
Chek your box.
Red: wrong
Green: orret
y
t tr
nex
ano
.
on
esti
qu
ther
re-enter
no
DO NOT PRESS THE
\E" (enter)
BUTTON.
no
TWO-LETTER CODES
Choie 1: AA
Choie 2: AB
Choie 3: AC
Choie 4: AD
Choie 5: BA
Choie 6: BB
Choie 7: BC
Choie 8: BD
Choie 9: CA
Choie 10: CB
previous question: CC
another question: CD
enter: E
midterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 6 2007, 11:00 pm
Question 1, chap 35, sect 1.
part 1 of 2
10 points
Consider a monochromatic light ray which
has a wavelength λvac in vacuum. The light
ray is incident normally from medium # 1
with an index of refraction n1 into a thin
film with an index of refraction n2 and a
thickness t. The ray is perpendicular to the
surface. (For clarity the ray is drawn with
a finite incident angle.) This film is coated
on a plastic medium which has an index of
refraction n3 . Assume n2 > n1 > 1, and
n2 > n3 . Denote
(i) φrefl to be that part of the phase angle
difference between the exit rays # 2 and # 1
if the path difference between ray 1 and ray 2
is neglected.
(ii) φpath to be that part of the phase angle
difference between the exit rays # 2 and # 1
due to the path difference alone.
Caution: n1 > 1.
The quantities φrefl and φpath are respectively
given
by
#1 #2
4
2 π n2 t
n3 λvac
2 π n2 t
8. π ,
n3 λvac
2 π n2 t
9. π ,
n1 λvac
2 π n2 t
10. 0 ,
n1 λvac
Explanation:
7. 0 ,
ref
φref = |φref
1 − φ2 | = |π − 0| = π.
φpath =
2t
4 π t n2
2π =
.
λ2
λvac
Question 2, chap 35, sect 1.
part 2 of 2
10 points
Given n1 = 1.2, n2 = 1.6, and n3 = 1.5.
The incident wave in the vacuum has a wavelength of 520 nm. If the reflected ray corresponds to a minimum, find the least nonzero
thickness of the film.(Hint: This question uses
the part 1 as the intermediate step).
Correct answer: 162.5.
Explanation:
n1
t
φ = φpath + φrefl .
The minima sequence is given by:
n2
π, 3 π, ... = 2 π
n3
1. π ,
2. π ,
3. 0 ,
4. 0 ,
5. 0 ,
6. π ,
4 π n2 t
n1 λvac
4 π n2 t
n3 λvac
4 π n2 t
n3 λvac
4 π n2 t
n1 λvac
4 π n2 t
λvac
4 π n2 t
correct
λvac
2t
+ |π − 0|.
λ2
For the least nonzero thickness, on the right
hand side we take
3π = 2π
2t
+ |π − 0|.
λ2
This leads to
2π = 2π
or
t=
2t
λ2
λvac
λ2
.
=
2
2 n2
Question 3, chap 34, sect 3.
part 1 of 1
10 points
midterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 6 2007, 11:00 pm
A jewel thief decides to hide a stolen diamond by placing it at the bottom of a crystalclear fountain. He places a circular piece of
wood on the surface of the water and anchors
it directly above the diamond at the bottom
of the fountain.
d
1.02 m
If the fountain is 1.02 m deep, find the
minimum diameter, d of the piece of wood
that would prevent the diamond from being
seen from outside the water. The index of
refraction of air is 1, that of water is 1.333,
and that of diamond is 2.4.
Correct answer: 2.31447.
Explanation:
Basic Concept:
sin θc =
nr
ni
Given:
ni = 1.333
nr = 1.00
∆y = 1.02 m
Solution:
nr
θc = sin
n
i 1
= sin−1
1.333
◦
= 48.6066
−1
tan θc =
d
2
∆y
d
=
2∆y
d = 2∆y(tan θc )
= 2(1.02 m)(tan 48.6066◦ )
= 2.31447 m
5
Question 4, chap 35, sect 6.
part 1 of 1
10 points
NASA puts a spy satellite into circular orbit
around Earth at a radius of 1.3 × 107 m in a
top secret mission to determine the number of
students in a school band.
Assume: A green filter is used (wavelength
574 nm), and that the band is in formation in
which everyone is separated by 1.8 m.
How large of a diameter must the lens of the
spy satellite be in order to allow an accurate
count?
Consider ≈ 6 × 106 m . to be the approximate radius of the Earth.
Correct answer: 2.72331.
Explanation:
Let : Re = 6 × 106 m = 6 Mm ,
Rs = 1.3 × 107 m = 13 Mm ,
Rd = 7 × 106 m = 7 Mm ,
ℓ = 1.8 m , and
λ = 574 nm .
Basic Concepts: The Rayleigh criterion
1.22 λ
is θ =
, where D is the lens diameter
D
ℓ
.
and the angle θ ≡
Rs
Solution: Using the Rayleigh criteria we get
1
θ
Re
= 1.22 λ
ℓ
= 1.22 (5.74 × 10−7 m)
(7 × 106 m)
×
(1.8 m)
= 2.72331 m .
D = 1.22 λ
Question 5, chap 34, sect 5.
part 1 of 2
10 points
midterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 6 2007, 11:00 pm
Given: A real object is located at “p1 =
11
f ” to the left of a convergent lens with a
6
focal length f as shown in the figure below.
f
f
Basic Concepts:
1 1
1
h′
q
+ =
m=
=−
p q
f
h
p
Converging Lens
f >0
∞ >p> f
f >p> 0
f <q< ∞
−∞ < q < 0
Diverging Lens
11
0
6
The image distance q1 to the right of the
lens is
6
∞ >p> 0
0 > m > −∞
∞ >m> 1
f <q< 0
0>f
0 <m< 1
(× f )
1. q1 =
2. q1 =
3. q1 =
4. q1 =
5. q1 =
6. q1 =
7. q1 =
8. q1 =
9. q1 =
10. q1 =
7
f.
2
9
f.
4
15
f.
4
15
f.
7
11
f.
6
13
f.
5
8
f.
3
11
f . correct
5
10
f.
3
5
f.
2
Convergent (concave) lenses f > 0 have
real images q > 0 when the object ∞ > p > f .
f
11
6
Basic Concepts:
(× f )
f
0
11
5
1 1
1
+ = .
p q
f
Solution:
1
1
1
=− +
q1
p1 f
1
6
+
=−
11 f
f
11 − 6
=
11 f
5
=
11 f
11
q1 =
f .
5
Question 6, chap 34, sect 5.
part 2 of 2
10 points
Explanation:
A second convergent lens with the same
focal length f is placed behind the first lens,
as shown in the figure below (the first lens has
the lighter image).
midterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 6 2007, 11:00 pm
7
11
and from Part 2, we have q1 =
f.
5
Solution: q1 > d and for the second lens
f
f
f
f
11
4
0
6
5
The second image location q2′ , measured
with respect to the second lens, is
(× f )
1. q2′ =
2. q2′ =
3. q2′ =
4. q2′ =
5. q2′ =
6. q2′ =
7. q2′ =
8. q2′ =
9. q2′ =
10. q2′ =
1
f.
2
5
f.
7
2
f.
3
8
f.
13
9
f.
16
5
f.
8
7
f . correct
12
8
f.
11
9
f.
14
4
f.
7
The final image due to the second lens is real
and to the right of the second lens. As can
be seen from the figure, the final image due to
the second lens relative to the first lens is
7
f
5
4
7
f+
f
5
12
83
=
f .
60
q2′ =
f
83
f
60
(× f )
11
6
Note: Since the object for the second lens
is behind the lens it is virtual and therefore
p2 < 0.
Using the lens equation for the second lens,
we have
1
1
1
=− +
q2
p2 f
1
1
+
=
q1 − d f
1
1
+
=
4
11
f
f− f
5
5
7
5
+
=
7f
7f
5+7
=
7f
12
=
7f
7
f.
q2 =
12
Explanation:
f
p2 = −q1 + d
11 4
f
= − +
5
5
7
=− f.
5
Question 7, chap 35, sect 3.
part 1 of 1
10 points
0
4 7 11
5 12 5
Given: Distance between lenses is d =
4
f,
5
A screen is illuminated by 530 nm light as
shown in the figure below.
The distance from the slits to the screen is
3.8 m .
midterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 6 2007, 11:00 pm
2 φ
I = I0 cos
, with φ = k δ .
2
8
(1)
68%
Let :
S1
θ
viewing
screen
0.44 mm
y
S2
3.8 m
Solution: Using Eq. 1, we have
Figure: Not drawn to scale.
Find the minimum positive phase angular
I
value φ such that
= 68% , where I0 is the
I0
intensity at the central maximum and I is the
intensity at the position y on the screen.
Correct answer: 68.8998.
φ = 2 arccos
= 2 arccos
π
π/2
2π
1
or
3 π/2
2π
360◦
φ
0
−1
=λ
λ = 530 nm
0◦
90◦
180◦
270◦
360◦
d y
φ
δ
d sin θ
d
y
≈
≡ =
= p
2
2
2π
λ
λ
λ L +y
λ L
Double slit interference
the phase angle difference φ
maxima: φ = 2 π , 4 π , 6 , 8 π · · · ,
minimum: φ = π , 3 π , 5 , 7 π · · · .
2
I∝E =
4 E02 cos2
φ
φ
2
cos ω t +
2
2
h√
I
I0
#
0.68
i
Note: The difference in path length for the
rays can be found as follows.
δ
φ
Using
≡
, and Eq. 1, we have
λ
2π
δ=λ
δ
"r
= 2 arccos(0.824621)
= 2 (0.601264 rad)
= 1.20253 rad
= 68.8998 ◦ .
Explanation:
Basic Concepts:
0
I(φ)
I
=
I0
I(0)
= 68%
= 0.68
φ
2π

r 

I 




 2 arccos
I 
0




2π




= (5.3 × 10−7 m)
(
√
)
2 arccos 0.68
×
2π
(1.20253)
−7
= (5.3 × 10 m)
2π
= (0.00044 m){0.000230536 rad}
= 1.01436 × 10−7 m
= 101.436 nm .
Note: The angle θ can be found as follows:
d
S2
Let :
y
viewing
screen
S1
midterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 6 2007, 11:00 pm
9
(5.3 × 10−7 m) (1.20253)
P
= L tan arcsin
6 S2 Q S1 ≈ 90◦
2 π (0.00044 m)
= L tan(arcsin[0.000230536])
y
r1
= (3.8 m) tan(0.000230536 rad)
−1
an
L
r2 θ = t
= (3.8 m) tan(0.0132087 ◦ )
O
= 0.000876037 m
Q
δ ≈ d sin θ ≈ r2 − r1
= 0.876037 mm .
L
I
= 0.68
I0
λ = 530 nm = 5.3 × 10−7 m
L = 3.8 m , and
d = 0.44 mm = 0.00044 m .
Solution: Using Eq. 1 and
δ
d sin θ
φ
≡ =
, we have
2π
λ
λ
λφ
θ = arcsin
2πd
(
"r #)
λ
I
= arcsin
(2)
2 arccos
2πd
I0
(5.3 × 10−7 m)
= arcsin
2 π (0.00044 m)
h√
i
× 2 arccos 0.68
(5.3 × 10−7 m) (1.20253)
= arcsin
2 π (0.00044 m)
= arcsin(0.000230536)
= 0.000230536 rad
= 0.0132087 ◦ .
y
Note: Since tan θ = , and using Eqs. 1
L
and 2, we have
y = L tan θ
= L tan arcsin
(
λ
arccos
πd
"r
(5.3 × 10−7 m)
= L tan arcsin
2 π (0.00044 m)
h√
i
× 2 arccos 0.68
I
I0
#)!
Question 8, chap 34, sect 2.
part 1 of 1
10 points
Consider the case in which light ray A is
incident on mirror 1, as shown in the figure.
The reflected ray is incident on mirror 2 and
subsequently reflected as ray B. Let the angle of incidence (with respect to the normal)
on mirror 1 be θA = 48 ◦ and the point of
incidence be located 20 cm from the edge of
contact between the two mirrors.
mirror 2
B
θB
A
θA
90◦
mirror 1
x
What is the angle of the reflection of ray B
(with respect to the normal) on mirror 2?
Correct answer: 42.
Explanation:
If the angle of the incident ray A is θA , the
angle of reflection must also be θA . Since the
mirrors are perpendicular to each other, angle
θB is equal to 90◦ − θA
θB = 90◦ − θA
= 90◦ − 48◦
= 42◦ .
midterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 6 2007, 11:00 pm
10
is located at the first single-slit diffraction
minimum.
Question 9, chap 35, sect 5.
part 1 of 1
10 points
y
Consider the setup of a single slit experiment shown in the figure.
The first minimum for 410 nm light is at y1 .
The first minimum for 540 nm light is at y2 .
y
S1
θ
L
y2
?
y1
Correct answer: 1.31707.
What is the ratio
Explanation:
The first minimum is at
y=
λL
a
For the first light,
S2
L
d
Determine the ratio ; i.e., the slit separaa
tion d compared to the slit width a.
1.
2.
y1 =
λ1 L
a
For the second light,
y2 =
3.
4.
λ2 L
a
Therefore
λ2
y2
=
y1
λ1
(540 nm)
=
(410 nm)
= 1.31707 .
5.
6.
7.
8.
9.
Question 10, chap 35, sect 5.
part 1 of 1
10 points
Hint: Use a small angle approximation; e.g.,
sin θ ≈ tan θ ≈ θ and cos θ ≈ 1 .
Consider the setup of double-slit experiment
in the schematic drawing below.
Note: As can be seen in the figure below,
one of the double-slit interference maxima
viewing screen
S2
viewing
screen
d
θ
a
a
S1
10.
d
a
d
a
d
a
d
a
d
a
d
a
d
a
d
a
d
a
d
a
=
11
2
=5
=4
=2
=
3
2
= 3 correct
=
5
2
=1
7
2
9
=
2
=
Explanation:
At y there is a minimum for single-slit
diffraction and a maxima for double-slit interference, as noted in the question.
midterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 6 2007, 11:00 pm
The first minimum for single-slit diffraction
occurs when
λ
sin θ = ,
(1)
a
and the maxima for double-slit interference
occur when
λ
(2)
sin θ = (m) .
d
The first diffraction minimum for single-slit
diffraction and the third double-slit interference maximum (m = 3) occur at the same
position y, as seen in the figure below.
11
Question 11, chap 35, sect 4.
part 1 of 1
10 points
Using a grating with a spacing of 0.0004 cm,
a line (which looks like red light) appears
21 cm from the central line on a screen. The
screen is 1.2 m from the grating.
What is the wavelength of the light?
Correct answer: 689.521.
Explanation:
Given:
d = 4 × 10−6 m
L = 1.2 m
x = 21 cm
λ =?
3
For diffraction gratings,
0
m
Figure: The dashed curve on the
left of the screen is due to single
slit interference. The dashed curve
on the right of the screen is due to
double slit interference. The screen
position is zero amplitude and the
positive direction is reflected on either side of the screen.
Since the single-slit diffraction minimum
masks the third double-slit interference maxima, one must estimate where the third
double-slit maxima occurs using the spacing
between the double-slit interference pattern
shown on the right-hand side of the viewing
screen, as seen in the figure above.
Using Eq. 1 and 2, we have
sin θ
1
1
= = (m)
λ
a
d
d
= (3)
a
= 3.
λ = d sin θ
x
=d √
x2 + L 2
= (0.0004 cm)
"
×
p
(21 cm)
(21 cm)2 + (1.2 m)2
!#
= (6.89521 × 10−7 m) (1 × 109 nm/m)
= 689.521 nm .
Note:
sin θ = √
tan θ =
x2
x
= 0.17238 ,
+ L2
and
x
= 0.175 .
L
If sin θ ≈ tan θ the small angle approximation
can be used. Then we would have
λ = d sin θ
≈ d tan θ
x
≈ (0.0004 cm)
L
0.21 m
≈ (0.0004 cm)
1.2 m
≈ 700 nm ,
midterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 6 2007, 11:00 pm
with a fractional error of
λsin − λtan
(689.521 nm) − (700 nm)
=
λsin
(700 nm)
= −0.015197
= −1.5 % .
Dimensional analysis of λ
cm · cm
m
1m
102 cm
2
109 nm
1m
= nm
Question 12, chap 34, sect 3.
part 1 of 1
10 points
A light ray passes through substances 1, 2,
and 3, as shown.
The indices of refraction for these three
substances are n1 , n2 , and n3 , respectively.
Ray segments in 1 and 3 are parallel.
1
n1
n2
2
3
since θ1 = θ3 , sin θ1 = sin θ3 and n1 = n3 .
Since sin θ1 = sin θ3 > sin θ2 , n2 > n1 = n3 .
So, we can only conclude that n3 must be
the same as n1 .
Question 13, chap 34, sect 4.
part 1 of 1
10 points
A spherical Christmas tree ornament is
3.6 cm in diameter.
What is the magnification of the image of
an object placed 11.2 cm away from the ornament?
Correct answer: 0.0743802.
Explanation:
Given :
p = 11.2 cm and
−3.6 cm
R=
= −1.8 cm .
2
Using the mirror equation,
θ1
θ2
d
x
12
n3
One can conclude that
1. all three indices must be the same.
2. n1 must be equal to 1.00.
3. n2 must be the same as n3 .
4. None of the six other options listed are
correct.
5. n3 must be the same as n1 . correct
6. n2 must be less than n1 .
7. n2 must be less than n3 .
Explanation:
n1 sin θ1 = n2 sin θ2 = n3 sin θ3 ,
2
1 1
1
=
= +
f
R
p q
1
2
1
2p − R
= − =
q
R p
Rp
Rp
q=
,
2P − R
and the magnification of the image is
M =−
q
p
R
2p − R
(−1.8 cm)
=−
2(11.2 cm) − (−1.8 cm)
=−
= 0.0743802 .
Question 14, chap 34, sect 6.
part 1 of 1
10 points
A biology student uses a simple magnifier
to examine the structural features of the wing
of an insect. The wing is held 2.28 cm in front
of the lens, and the image is formed 43 cm
from the eye.
What is the focal length of the lens?
midterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 6 2007, 11:00 pm
Correct answer: 2.40766.
2
9
2
7. λ =
9
2
8. λ =
9
6. λ =
Explanation:
Given :
p = 2.28 cm and
q = −43 cm .
From the thin lens equation,
pq
f=
p+q
(2.28 cm) (−43 cm)
=
2.28 cm + (−43 cm)
dy
L
Ly
d
dL
y
9. None of these.
1 dL
5 y
10. λ =
= 2.40766 cm .
Explanation:
r1
an
r2 θ = t
S1
d
S2
−1
viewing
screen
L
Let L be the distance from the source to the
screen and h be the distance from the source
to the mirror.
Using the small angle approximation (θ =
sin θ = tan θ) , if the distance from the central
bright region to the fifth bright fringe is y,
what is the wave length of the light?
2. λ =
3. λ =
4. λ =
5. λ =
1
5
1
6
1
6
1
6
1
5
Ly
d
dy
L
dL
y
Ly
d
dy
correct
L
y
O
δ ≈ d sin θ ≈ r2 − r1
L
r1
S1
θ
S2
1. λ =
L
Q
y
d
y
viewing
screen
A screen is illuminated by monochromatic
light whose wave length is λ, as shown in the
figure below.
The distance from the slits to the screen is
L.
P
S2 Q S1 ≈ 90◦
6
Question 15, chap 35, sect 3.
part 1 of 1
10 points
S1
13
−1
θ = t an
h
d=
2
θ
S2
y
r2
Q
6
δ
L
◦
0
≈9
r1
S1
r2 −
S2 Q
≈
nθ
h si
2
≈
The angle θ from the slits’ midpoint to the
y position on the screen is
θ = arctan
y
L
.
The wavelength of the light for the fifth
bright fringe (m = 5) is
d sin θ
m
n
y o
d
sin arctan
=
m
L
λ=
midterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 6 2007, 11:00 pm
dy
mL
1 dy
,
=
5 L
Should the mirror be concave or convex?
=
1. Convex
2. Concave correct
after using the small angle approximation.
Question 16, chap 34, sect 4.
part 1 of 2
10 points
A dentist wants a small mirror that produces an upright image with a magnification
of 4 when the mirror is located 3 cm from a
tooth.
What should be the radius of curvature of
the mirror?
Correct answer: 8.
Explanation:
The lateral magnification of the mirror is
given by
s′
m=− ,
s
so
s′ = −m s .
From the mirror equation, we have
r=
=
=
=
=
Explanation:
The mirror must be concave because a convex mirror always produces a diminished virtual image.
Question 18, chap 34, sect 5.
part 1 of 2
10 points
A person looks at a gem using a converging
lens with a focal length of 13.4 cm. The lens
forms a virtual image 22.6 cm from the lens.
a) Find the linear magnification.
Correct answer: 2.68657.
Explanation:
Basic Concept:
m = 4 , and
s = 3 cm .
2
1
1
s′ + s
1
= = + ′ = ′ ,
f
r
s s
s s
14
so
2 s s′
s + s′
2 s (−m s)
s + (−ms)
−2 m s
1−m
−2 (4) (3 cm)
1−4
8 cm .
Question 17, chap 34, sect 4.
part 2 of 2
10 points
1
1 1
= +
f
p q
Given:
q = −22.6 cm
f = 13.4 cm
Solution:
1
1 1
= −
p
f
q
−1
1 1
−
p=
f
q
−1
1
1
−
=
13.4 cm −22.6 cm
= 8.41222 cm
−q
p
−(−22.6 cm)
=
8.41222 cm
= 2.68657
M=
Question 19, chap 34, sect 5.
part 2 of 2
10 points
midterm 04 – JYOTHINDRAN, VISHNU – Due: Dec 6 2007, 11:00 pm
b) Describe the image from the previous
problem.
1. real, inverted, smaller
2. virtual, upright, larger correct
3. virtual, inverted, larger
15
4. Yes for the grating with the blue light
bent less than the red light and no for the
prism.
5. Yes for both with blue light bend less
than the red light when passing through the
grating and blue light bent less than the red
light while passing through the prism.
4. virtual, upright, smaller
5. real, upright, larger
6. real, upright, smaller
6. Yes for both with the blue light bent less
than the red light when passing through the
grating and blue light bent more than the
red light when passing through the prism.
correct
7. real, inverted, larger
8. virtual, inverted, smaller
Explanation:
q < 0 so the image is virtual. M > 0 so
the image is upright. |M | > 1 so the image is
magnified.
Question 20, chap 34, sect 5.
part 1 of 1
10 points
In an attempt to separate white light into
it’s constituent colors, a beam if white light
from an incandescent bulb is allowed to pass
through either a diffraction grating or a refractive prism.
Do primary bands of color emerge for each
case, and if so, what is the relationship between the light’s color and the degree to which
it bends?
1. Yes for both, with the blue light bent more
than the red light when passing through the
grating and with blue light bent less than the
red light while passing through the prism.
2. Yes for both, with the blue light bent
more than the red light when passing through
the grating and blue light more than the red
light when passing through the presm.
3. Yes for the prism, with the blue light
bent less than the red light, and no for the
grating.
7. Yes for the grating, with the blue light
bent more than the red light, and no for the
prism.
8. Yes for the prism, with the blue light
bent more than the red light and no for the
grating.
Explanation:
Blue light has a shorter wavelength than
red light does. The proportion by which light
will be bent by a diffraction grating will be
proportional to λd , where λ is the wavelength
of the light and d is the spacing of the grating. Therefore, red light is bent more in a
diffraction grating than blue light.
Meanwhile, however, the bending of light
in a prism is caused by the slowing down of
light by the medium through which it travels.
Most materials, in the visible light range will
tend to slow down high frequency light more
than they will slow down low frequencly light.
Since the bending effect is caused by the slowing effect, this means that blue light will be
bent more than red light.