# exam3sol

```PHY 2049 Spring 2013 Exam 3 Solutions Exam 3 solutions Problem 1 The electric field of a plane electromagnetic wave is described by Em i sin(kz + ω t). Which expression describes the magnetic field associated to this electromagnetic wave? (1) −(Em/c) j sin(kz + ω t) (2) (Em/c) i sin(kz + ω t) (3) (Em/c) j sin(kz + ω t) (4) −c Em j sin(kz + ω t) (5) c Em k sin(kz + ω t) The direction of the wave is in the –k direction. This direction should also be given by the Poynting vector: 1
S=
E × B m µ0 m
This implies the direction of Bm is –j The magnitude of Bm is given by Em
=c
Bm
Em
Bm =
c
Problem 2 A beam of initially unpolarized light is sent along the z−axis into a stack of three polarizing sheets placed perpendicularly to the z−axis, as shown in the figure. The angles θ1, θ2, and θ3 of the polarizing directions are measured counterclockwise from the positive direction of the y−axis. What percentage of the light's initial intensity is transmitted by the system when θ1=30°, θ2=−30°, and θ3=0? (1) 9% (2) 38% (3) 28% (4) 42% (5) 22% PHY 2049 Spring 2013 Exam 3 solutions The intensity is given by I
I = 0 cos 2 (θ 2 − θ1 ) cos 2 (θ 3 − θ 2 )
2
I 1
2

2

⇒ = cos ( 60 ) cos ( 30 ) = 9%
I0 2
Problem 3 A light ray traveling in a material with an index of refraction of n1 is incident onto a stack of two other materials with indices of refraction n2 and n3, respectively, as shown in the diagram. What is the minimum angle of incidence θ1 such that the light is totally reflected by the stack if n1=1.48, n2=1.50, and n3=1.43 ? (1) 75° (2) 72.5° (3) 80.5° (4) 18° (5) 90° We want to find θ3=90° : Refraction gives us n1 sin θ1 = n2 sin θ 2 = n3 sin θ 3
sin θ1 =
n3
= 0.966
n1
⇒ θ1 = 75
Note that there cannot be any internal reflection between the first and second materials as one needs to go to a lower index of refraction Problem 4 What is the solar radiation pressure in Pascals (N/m2) on a dark patch of the Earth's surface (perpendicular to the direction of the overhead sunlight) if the total power produced by the sun is 3.8×1026 W and the distance between the sun and the Earth is 1.5×1011 m? (1) 4.5×10−6 (2) 1.3×1018 (3) 1300 (4) 9×10−17 (5) 1.7×104 PHY 2049 Spring 2013 Exam 3 solutions The radiation pressure for totally absorbed light is: I
pr =
c
P
I= 2
4π d
pr = 4.5 × 10 −6 N/m 2
Problem 5 How long does it take for the current in a resistor to reach one−third of its maximum value of 4 mA after it is connected in series with a 0.3 H inductor to a 12 V EMF source? (1) 4 ×10−5 s (2) 3 ×10−5 s (3) 1 ×10−4 s (4) 2 ×10−5 s (5) 7 ×10−5 s The time constant for LR circuits is τ = L / R The solution for charging the circuit is i ( t ) = im (1− e−t /τ )
ε
ε
where im =
⇒ R = = 3kΩ
R
im
To find the time 1
im = im 1− e−t /τ
3
2
= e−t /τ
3
3 L 3
0.3
3
⇒ t = τ ln = ln =
ln = 4 × 10 −6 s
2 R 2 3000 2
Problem 6 An LC circuit has a capacitance of 20μF and an inductance of 10 mH. At time t=0 the charge on the capacitor is 27 μC and the current is 80 mA. The maximum possible charge is: (1) 45 μC (2) 27 μC (3) 100 μC (4) 63 μC (5) 36 μC The energy in an LC circuit is conserved. 1 2 q2
U = Li +
= 5 × 10 −5 J
2
2C
2
q
U = max
2C
⇒ qmax = 45 µF
(
)
PHY 2049 Spring 2013 Exam 3 solutions Problem 7 A series RLC circuit is driven by a sinusoidally−varying EMF source. The current leads the EMF by 30°. What can be concluded about the driving frequency ω? (1) ω < 1/√LC (2) ω > 1/√LC (3) ω = 1/√LC (4) ω = 0 (5) ω = ∞ The expressions for the EMF and current are ε ( t ) = ε m sin (ω t )
i ( t ) = im sin (ω t − φ )
Where a negative phase means that the current leads the EMF, and is given by: 1
ωL −
ω C tan φ =
R
So we must have ω < 1/√LC Problem 8 A 5μF capacitor is connected to an AC generator with ε = 25.0 V. What is the amplitude of the resulting alternating current if the frequency of the EMF in cycles/sec is f=7 kHz. (1) 5.5 A (2) 0.9 A (3) 1.3×10−4 A (4) 115 A (5) 5×10−3 A We can solve this using the loop rule ε ( t ) = ε m sin (ω t )
q
=0
C
q = Cε m sin (ω t )
ε−
dq
= Cωε m cos (ω t )
dt
⇒ im = Cω = 2π fCε m = 5.5A
i=
Problem 9 Suppose the magnetic field in a region of space has the form B(x,y,z) = 5 x i + 6 j + f(x,z) k, where f(x,z) is an unknown function. Which of the following choices for f(x,z) is allowed by Maxwell's equations? (1) f(x,z) = −5 z (2) f(x,z) = −5 x (3) f(x,z) = −11 z (4) f(x,z) = +5 z (5) f(x,z) = +5 x PHY 2049 Spring 2013 Exam 3 solutions Use Gauss's law for magnetic fields on a cube of side length L with one corner at the origin and edges along the positive x, y and z axes. The flux out of the cube through the face at x=0 is 0, and the flux out of the face at x=L is 5 L3. The outward flux through the face at y=0 is -­‐6 L2, and the outward flux through the face at y = L is +6 L2. The outward flux through the face at z=0 is: L
−L ∫ d xf (x,0) , 0
and the outward flux through the face at z=L is: L
+L ∫ d xf (x, L) 0
L
Hence the total outward flux is Φ B = 5L3 + L ∫ d x[ f (x, L) − f (x,0)] 0
This must vanish by Gauss's law for magnetic fields. The actual results for the five choices are: (1) f(x,z) = -­‐5z àΦB = 0. (2) f(x,z) = -­‐5x à ΦB = +5 L3 (3) f(x,z) = -­‐11z à ΦB = -­‐6 L3 (4) f(x,z) = +5z à ΦB = +10 L3 (5) f(x,z) = +5x à ΦB = +5 L3 Problem 10 An air−filled capacitor is formed from two parallel, circular plates of radius 5 cm which are separated by a distance of 2 mm. The capacitor is connected in series with a 5 Ω resistor and charged up from zero by a 12 volt battery. What is the magnitude of the magnetic field (in μT) inside the capacitor at a distance of 2 cm from the center, when the charge on the capacitor has reached half of its maximum value? (1) 2 (2) 5 (3) 12 (4) .2 (5) .4 Use the Ampere-­‐Maxwell law choosing the surface to be a circle of radius r, parallel to the two plates and centered on the same axis. By symmetry the magnetic field is circumferential and depends only upon time and the distance r from the central axis. Hence the left hand side of the Ampere-­‐Maxwell law is 2π r B(t,r). The right hand side is μ0 ε0 π r2 dE(t)/dt, where the electric field inside the capacitor is E(t) = Q(t)/ε0 A and A is the area of one of the plates. The charge on the capacitor is Q(t) = C ε [1 -­‐ e-­‐t/RC]. Taking the derivative and assuming the capacitor is half charged gives dQ/dt = ε /2R. Hence the right hand side of the Ampere-­‐Maxwell law is μ0 π r2 ε /(2 A R). Therefore, the magnetic field is B = μ0 r ε /(4 A R). PHY 2049 Spring 2013 Exam 3 solutions Problem 11 The famous 21 cm radiation of radio astronomy arises from flipping the spin of a Hydrogen atom electron in the presence of the magnetic field it perceives from the Hydrogen atom nucleus. The energy of this transition is about 5.9 μeV. What must be the strength of the magnetic field (in Tesla) the electron perceives? (1) 0.05 (2) 0.10 (3) 3 × 1017 (4) 6 × 1017 (5) 8 × 108 Recall that the energy of a magnetic dipole μ in the presence of an external magnetic field is U = -­‐ μ  B. So the energy different between being aligned with the magnetic field and being anti-­‐aligned is Δ U = 2 μ B. Recall also that the magnetic dipole moment of an electron is one Bohr magneton μB. Problem 12 Recall the in−class demonstration in which liquid oxygen (LOX) was dribbled into the horizontal region between the two poles of a magnet. Which of the following statements explains why the LOX remained suspended between the poles of the magnet? (1) LOX is paramagnetic and magnetic dipoles stick together (2) LOX is paramagnetic (3) LOX is diamagnetic and magnetic dipoles stick together (4) LOX is diamagnetic (5) Magnetic dipoles stick together Recall that paramagnetic materials develop dipoles in the same direction as the applied magnetic field. Recall also that aligned dipoles stick together. Problem 13 What kind of image (real = R or virtual = V) and orientation (inverted = I or not inverted = NI) is produced by a spherical cocave mirror? (1) Insufficient information (2) R, I (3) R, NI (4) V, I (5) V, NI PHY 2049 Spring 2013 Exam 3 solutions Recall Table 34-­‐1, which was derived in class. For a spherical concave mirror the type and orientation of the image depends upon the location of the object. If the object is inside the focal point then the image is virtual and not inverted; if the object is outside the focal point then the image is real and inverted. Because you were not told the object location there is insufficient information to determine the image location and orientation. Problem 14 Consider a spherical glass (n = 1.5) paperweight of diameter 5 cm that contains a grain of rice which appears (to an observer in air) to lie 1 cm from the edge of the glass. How far inside the glass is the grain actually located? (1) 1.25 cm (2) 1.00 cm (3) 0.57 cm (4) 1.50 cm (5) 1.75 cm This is almost identical to the sample problem on page 935. Because the paperweight is concave from the perspective of the object, its radius of curvature is r = -­‐2.5 cm. Because the image is formed inside the paperweight we have i = -­‐1 cm. Hence the object position p is found from the relation 1.5/p -­‐1 = -­‐.5/-­‐2.5. Problem 15 A diverging lens is formed by joining together two identical concave surfaces whose radii of curvature have magnitude 10 cm. If the focal length of the lens is −15 cm, what is the index of refraction of the material in the lens? (1) 4/3 (2) 5/3 (3) 1 (4) 3/2 (5) 5/4 From the lens maker's equation we have -­‐1/15 = (n-­‐1) [-­‐1/10 -­‐ 1/10]. Hence n = 1 + 5/15 = 4/3. Problem 16 An object is placed 5 cm from the first of a 2−lens system. The first lens has a focal distance of 10 cm, the second lens is 20 cm from the first and has a focal length of 20 cm. How far from the second lens does the image form? (1) 60 cm (2) 50 cm (3) 40 cm (4) 30 cm (5) 20 cm Use the 2-­‐step procedure explained in the text and in class. The first image obeys 1/5 + 1/i1 = 1/10. Hence we have i1 = -­‐10 cm, which makes p2 = 20 cm + 10 cm = 30 cm. The second image obeys 1/30 + 1/i2 = 1/20. Hence i2 = 60 cm. PHY 2049 Spring 2013 Exam 3 solutions Problem 17 Two narrow slits are separated by a distance of 5 microns. Monochromatic light of wavelength 500 nm passes through the slits and goes on to hit a screen 2 meters behind them. What is the distance between the 1st intensity minimum and the 3rd maximum (counting the central peak as the 0th maximum)? (1) 53 cm (2) 50 cm (3) 73 cm (4) 63 cm (5) 10 cm The first intensity minimum obeys sin(θ1) = λ/2d, hence it is located at y1 = 2 m × tan(θ1) ≈ 0.1001 m above the central maximum. The 3rd intensity maximum obeys sin(θ3) = 3 λ/d, hence it is located at y2 =2 m × tan(θ2) ≈ 0.6290 m above the central maximum. The distance between these two features is Δ y = y2 -­‐ y1 ≈ 0.5289 m. Note that the small angle approximation is not valid for the 3rd maximum. Problem 18 Light is normally incident from a region with index of refraction n1 = 1.40 to a layer of thickness 210 nm and index of refraction n2 = 1.46 which is followed by a region of index of refraction n3 = 1.75. Find the wavelength in air, in the visible range (400 nm to 700 nm), which results in the reflected wave being an interference minimum? (1) 409 nm (2) 420 nm (3) 613 nm (4) 588 nm (5) No visible light does this Ray 1 gets a phase shift upon reflection because n1 < n2. Ray 2 gets a phase shift upon reflection because n2 < n3. And ray 2 also acquires a phase of 2π × 2L n2/λ from propagating back and forth in the thin film. For a minimum we need the phase difference between the two rays to be a half integer multiple of 2π, hence 2L = (m + 1/2) λ/n2. The wavelength in air is therefore λ = 1.46 × 420 nm/(m + 1/2). The choice of m=1 gives a visible wavelength of λ = 408.8 nm. PHY 2049 Spring 2013 Exam 3 solutions Problem 19 Light of wavelength 587 nm in air is normally incident from a region with index of refraction n1 = 1.50 to a thin layer of index of refraction n2 = 1.34 which is followed by a region of index of refraction n3 = 1.42. What is the second thinnest layer that will result in the transmitted light being an interference minimum? (1) 329 nm (2) 219 nm (3) 438 nm (4) 294 nm (5) 310 nm Ray 4 gets two phase shifts because n2 < n3 and n2 < n1. It acquires an additional relative phase of 2π × 2L n2/λ from the extra back and forth propagation it executes in the thin film. For a minimum we need the relative phase to be a half integer multiple of 2 π, hence 2 L = (m + 1/2) λ/n2. The second thinnest layer corresponds to m = 1, or L = 3/4 × 587 nm/1.34 ≈ 329 nm. Problem 20 When monochromatic light of an unknown wavelength passes through a single slit of width 0.12 mm the first diffraction minimum is observed to be .5 cm away from the central maximum on a screen that is 1 m from the slit. What is the wavelength of the light? (1) 600 nm (2) 400 nm (3) 500 nm (4) 700 nm (5) 450 nm The first diffraction minimum obeys sin(θ ) = λ / a and tan(θ ) = y / D . Making the small angle approximation gives λ ≈ ay / D = 600 nm ```

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