Download Exam 3 Solutions

PHY 2049 Fall 2013 – Acosta, Woodard Exam 3 Solutions Exam 3 solutions Problem 1 A driven LRC circuit consists of a series connection of an inductor L = 1 mH, a resistor R = 10 Ω, and a capacitor C = 10 μF. Suppose it is driven by an EMF of E(t) = Em cos(ω t) . If ω max is the angular frequency at which the current amplitude is greatest, what is the ratio I(2ω max ) / I(ω max ) of the current amplitude at twice this angular frequency to it maximum value? (1) 0.55 (2) 0.90 (3) 0.75 (4) 0.40 (5) 0.25 Recall that the current at angular frequency ω is Em
I(ω ) =
1 2
2
R + (ω L −
)
ωC
Recall also that ω max = 1 / LC . At the two frequencies we have E
I(ω max ) = m
R
Em
I(2ω max ) =
9L
R2 +
4C
Hence their ratio is: I(2ω max )
1
2
=
=
≈ 0.55 I(ω mxa )
13
9L
1+
4CR 2
Problem 2 Suppose a primary circuit carries an AC current i1 and voltage V1. We transform this through a primary coil of 100 turns to a secondary coil of 5 turns. What are the amplitudes i2 and V2 of the current and voltage in the secondary circuit? (1) i2 = 20 i1 and V2 = V1/20 (2) i2 = 20 i1 and V2 = 20 V1 (3) i2 = i1/20 and V2 = 20 V1 (4) i2 = i1/20 and V2 = V1/20 (5) i2 = i1 and V2 = V1/20 PHY 2049 Fall 2013 – Acosta, Woodard Exam 3 solutions A transformer preserves the magnetic flux from one coil to the other so we have N1 i1 = N2 i2, which implies i2 = 20 i1. A transformer also preserves the power i1 V1 = i2 V2, so V2 = V1/20. Problem 3 A spherical, concave mirror is shown in the figure. The focal point F and the location of the object O are indicated. At what point will the image be located? (1) V (2) IV (3) III (4) II (5) I Recall from Table 34-­‐1 that the image from a concave mirror forms on the other side of the mirror when the object is inside the focal point. (You can also derive this by following the incoming rays which intersect the top of the object parallel to the central axis -­‐-­‐-­‐ and hence reflect back through the focal point -­‐-­‐-­‐ and an incoming ray which intersect's the top of the object through the focal point -­‐-­‐-­‐ and hence reflects back parallel to the central axis.) The only possible point for such an image is V. Problem 4 Suppose an object on the central axis of a spherical mirror is magnified by m = +4. Is the image on the same side (S) or opposite (O), is it real (R) or virtual (V), and is it inverted (I) or not inverted (NI)? (1) O, V, NI (2) S, V, I (3) S, R, I (4) O, R, NI (5) S, R, NI PHY 2049 Fall 2013 – Acosta, Woodard Exam 3 solutions Recall that the image distance i is positive when the image is on the same side as the human and negative when the image forms on the side opposite to the human. Recall also that m = -­‐i/p, so the image distance must be negative. This means the image is on the OPPOSITE side, it is VIRTUAL, and NOT INVERTED. Problem 5 Consider a spherical refracting surface has radius of curvature r = +30 cm. The object is at distance p = +70 cm in a medium with index of refraction n1 = 1.5. The observer is in air with n2 = 1.0. What is the image distance i, is the image real (R) or virtual (V), and is it on the same side (S) or opposite (O) as the object? (1) i = -­‐26 cm, V, S (2) i = -­‐32 cm, V, S (3) i = +32 cm, R, S (4) i = +26 cm, V, S (5) i = -­‐26 cm, R, O Because the curvature radius is positive, the surface must be convex with respect to the object. The equation n1 n2 (n2 − n1 )
+ =
r
p i
105
implies i = −
cm ≈ −26cm . 4
Because the image distance is negative, the image must be VIRTUAL and hence on the SAME side as the object. Problem 6 An object is located 40 cm from the first of two thin converging lenses of focal lengths f1 = 20 cm and f2 = 10 cm as shown in the figure. The lenses are separated by 30 cm. Where is the final image of this two-­‐lens system formed? PHY 2049 Fall 2013 – Acosta, Woodard Exam 3 solutions (1) 5.0 cm to the right of the second lens (2) 13.3 cm to the right of the second lens (3) infinitely far to the right of the second lens (4) 13.3 cm to the left of the second lens (5) 100 cm to the left of the second lens Recall that we make two uses of the equation 1 1 1
+ = p i f
In the first step we ignore the second mirror and find where the image of the first mirror forms. The equation gives i1 = 40 cm. We then compute the object distance to the second mirror as p2 = d − i1 = 30cm − 40cm = −10cm It makes no matter that this is negative! Just go ahead and plug it into the equation again to infer where the second image forms. The result is i2 = 5 cm, which means the image is 5 cm to the right of the second lens. It would also be REAL and NOT INVERTED, but we were not asked that. Problem 7 In the figure two light rays go through different paths by reflecting from the various flat surfaces shown. The light waves have a wavelength of 420 nm and are initially in phase. What are the smallest and second smallest values of L that will put the waves exactly out of phase as they emerge from the region? (1) 52 nm and 157 nm (2) 60 nm and 180 nm (3) 52 nm and 105 nm (4) 105 nm and 210 nm (5) 120 nm and 240 nm PHY 2049 Fall 2013 – Acosta, Woodard Exam 3 solutions The extra distance Ray 1 travels is 4L. Hence the phase difference between the two rays is Δφ = 2π / λ × 4L . For the two rays to be exactly out of phase 1
we must have Δφ = (N − )2π 2
1
1
Hence we have L = (N − )λ / 4 = (N − )105nm . 2
2
The smallest value is N=1, which gives L1 = 52.5 nm. The second smallest is N=2, which gives L2 = 157.5 nm. Problem 8 Monochromatic light of wavelength 500 nm passes through a pair of slits separated by 1 mm. What is the angle from the central maximum to the 1st minimum? (1) 0.015 degrees (2) 0.025 degrees (3) 0.075 degrees (4) 0.0055 degrees (5) 0.00035 degrees 1
Recall that the Nth double slit minimum occurs at sin(θ ) = (N − )λ / d . 2
The first minimum is N = 1 so we have θ = sin −1 (0.00025) ≈ 0.014 degrees Problem 9 Monochromatic light whose wavelength in air is 612 nm is normally incident from a region of n_1 =1.55 to a thin layer of n_2 = 1.60, which is followed by a region of n_3 = 1.33, as shown in the figure. What is the 3rd thinnest layer that maximizes the reflected light? (1) 478 nm (2) 669 nm (3) 861 nm (4) 96 nm (5) 287 nm PHY 2049 Fall 2013 – Acosta, Woodard Exam 3 solutions Because n1 < n2, Ray 1 experiences a phase shift; because n2 > n3, Ray 2 does not experience a phase shift. We want a maximum so the length L and the wavelength λ are related by 2L
1
+ 0 − = N + 0 2
λ / n2
1
Hence we have L = (N + ) × 191.25nm 2
The thinnest layer corresponds to N=0, so the 2nd thinnest is N=1 and the 3rd thinnest is N = 3. This gives L ≈ 478nm . Problem 10 Light is normally incident from a region of n1 = 1.32 to a layer of thickness 325 nm and n2 = 1.75, which is followed by a region of n3 = 1.39, as shown in the figure. What wavelength (in air) in the visible range ( 400nm < λ < 700nm ) will result in the refracted light being minimized? (1) 455 nm (2) 535 nm (3) 585 nm (4) 635 nm (5) 665 nm Because n2 > n3, Ray 4 experiences no phase shift on its first reflection. Because n2 > n1, Ray 4 also experiences no phase shift on its second reflection. We want a minimum so the length L and the wavelength λ are related by 2L
1
+ 0 + 0 = N + 2
λ / n2
1
1137.5nm
=
Hence we have λ = 2Ln2 ×
1
1
N+
N+
2
2
To get within the visible range we need N = 2, which gives λ = 455 nm. PHY 2049 Fall 2013 – Acosta, Woodard Exam 3 solutions Problem 11 Monochromatic light of wavelength 441 nm is incident on a narrow slit. On a screen 2 m away, the distance between the 2nd diffraction minimum and the central maximum is 1.50 cm. What is the angle of the second intensity minimum? (1) 0.43 degrees (2) 0.64 degrees (3) 1.3 degrees (4) 37 degrees (5) 0.0075 degrees From the geometry of the single slit we recall that tan(θ ) = y / D , where y = 0.015 m and D = 2 m. Hence we have θ = tan −1 (0.0075) ≈ 0.43 degrees Problem 12 The electric field of a plane electromagnetic wave propagating in vacuum is ! !
!
described by E = Em sin(ky + ω t) , where the amplitude Em = Em iˆ , the wavenumber is k, and the angular ! !frequency is ω. The associated magnetic field of the plane wave is described by B = Bm sin(ky + ω t) . In what direction does the amplitude of the !
associated magnetic field, Bm , point? (1) k̂ (2) î (3) ĵ (4) − ĵ (5) − k̂ The wave propagation direction is − ĵ from the ky+ωt term. The electric field direction is correctly orthogonal to that direction, î . The magnetic field 1
direction must be orthogonal to both those directions, such that S =
E × B µ0
gives the wave direction. This leads us to k̂ , since − ĵ = î × k̂ Problem 13 A beam of initially unpolarized light is sent into a stack of 3 polarizing sheets, where the polarizing direction of each sheet is rotated +60° with respect to the previous sheet. What fraction of the incident intensity is transmitted through the stack of 3 sheets? (1) 1/32 (2) 1/8 (3) 1/16 (4) 1/4 (5) 9/32 PHY 2049 Fall 2013 – Acosta, Woodard Exam 3 solutions The relative transmitted intensity through the 3 sheets is given by: 1
I 1
= cos 2 60! cos 2 60! =
32
I0 2
The factor ½ is the reduction in intensity of the initial unpolarized light through the first filter. ( )
( )
Problem 14 The magnetic dipole moment associated with an iron atom is 3.0 x 10-­‐23 J/T. Assume that all the atoms in an iron bar, which is 7.0 cm long and has a cross-­‐sectional area of 1.0 cm2, have their dipole moments aligned with an external magnetic field of 0.25 T. What energy must be provided to rotate the bar so that the all the atoms become oppositely aligned with the external field? The density of iron is 8.0 g/cm3 and its molar mass is 56 g/mol. Avogadro’s number is 6.022 x 1023. (1) 9.0 J (2) 4.5 J (3) 18 J (4) 1.3 J (5) 36 J The volume of the bar is V = 7 cm3, so the mass is M= ρV = 56 g. The number of iron atoms is therefore: 56g 6.022 × 10 23
N=
= 6.022 × 10 23 56 g/mol
mol
And the total magnetic dipole moment is µ = 6.022 × 10 23 3 × 10 −23 J/T = 18 J/T (
)(
)
The energy required to orient the bar so that the magnetic dipole is oppositely aligned with the external field from being aligned is given by the difference in the potential energy, where the potential energy of a magnetic !
dipole in a magnetic field is U = − µ ⋅ B . Thus W = ΔU = 2 µ B = 9 J Problem 15 The magnetic flux through the top face of the shown cube is +4.0 Wb, and that through the bottom face is -­‐1.0 Wb. What must be the sum of the magnetic flux passing through the 4 other sides of the cube? PHY 2049 Fall 2013 – Acosta, Woodard Exam 3 solutions (1) -­‐3.0 Wb (2) 3.0 Wb (3) -­‐5.0 Wb (4) 5.0 Wb (5) Insufficient information Gauss’ Law for magnetic fields says that the total magnetic flux through a closed surface must be zero (no monopoles). Therefor we must have: Φ tot = Φ top + Φ bottom + Φ sides = 0
Φ sides = − ( 4 − 1) = −3
Problem 16 A parallel-­‐plate capacitor, with circular plates of radius 4.0 cm, has a uniform electric field between the plates with a magnitude that increases linearly in time as 5t (in units of V/m). What is the magnitude of the magnetic field at a radius of 1 cm from the axis connecting the centers of the two plates? (1) 2.8 x 10-­‐19 T (2) 1.1 x 10-­‐18 T (3) 8.8 x 10-­‐16 T (4) 1.7 x 10-­‐20 T (5) 2.3 x 1015 T The magnetic field can be calculated from the changing electric flux using the Maxwell’s law of induction: dΦ
!∫ B ⋅ ds = ε 0 µ0 dt E
dE
2π rB = ε 0 µ0 π r 2
dt
5ε µ
B = 0 0 r = 2.8 × 10 −19 T
2
Problem 17 The peak amplitude of the electric field of an electromagnetic plane wave is given by Em=1000 V/m. What is the average intensity of the wave in units of W/m2? (1) 1.3 x 103 (2) 4.0 x 1011 (3) 1.2 x 1020 (4) 1.0 x 106 (5) 2.7 x 103 ( )
PHY 2049 Fall 2013 – Acosta, Woodard Exam 3 solutions The Poynting vector gives the intensity of an electromagnetic wave: 1
E × B µ0
When averaged over time, the magnitude of the intensity is: S=
1
Em Bm where Em and Bm are the peak electric and magnetic field 2 µ0
E
amplitudes. They are related by m = c . So Bm
1
Sav =
Em 2 = 1330 W/m 2 2 µ0 c
Sav =
Problem 18 A parallel-­‐plate capacitor, with circular plates of radius 2.0 cm, has a uniform electric field in the region between the plates and a magnetic field at a radius of 4.0 cm of 1.25 x 10-­‐5 T. What is the displacement current between the plates? (1) 2.5 A (2) 1.3 A (3) 3.1 x 10-­‐6 A (4) 2.8 x 1011 A (5) 0 A The magnetic field can be calculated from the changing electric flux, using the Maxwell’s law of induction: dΦ
!∫ B ⋅ ds = ε 0 µ0 dt E = µ0idisp where idisp is the displacement current. The radius of the plates is smaller than the radius to where the magnetic field is measured, so the entire displacement current contributes to the changing flux. So !∫ B ⋅ ds = µ0idisp
2π rB = µ0idisp
2π rB
= 2.5A
µ0
where r is taken to the radius where the magnetic field is known. idisp =
Problem 19 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. The light ray has an angle of incidence of Θ1=45° when it PHY 2049 Fall 2013 – Acosta, Woodard Exam 3 solutions enters the stack. What is the angle the light ray makes relative to the normal to the material surface when it enters the material with refractive index n3 if n1=1.0, n2=1.5, and n3=1.33? (1) 32° (2) 28° (3) 58° (4) 84° (5) 73° The law of refraction will yield (when combined with the fact that alternate interior angles are equal for two parallel lines cut by light ray): n1 sin θ1 = n2 sin θ 2 = n3 sin θ 3
⎛n
⎞
θ 3 = sin −1 ⎜ 1 sin θ1 ⎟ = 32!
⎝ n3
⎠
Due to some confusion on the wording, we’ll accept also the angle in n2, which is 28 degrees. Problem 20 A nighttime cave diver wants to explore a new horizontal cave that is located at a depth d=10m below the surface of the water. She will use a flashlight to signal her buddy in a boat on the surface of the water if she runs into trouble. What distance w could she go into the cave and still expect her light to be visible to her buddy in the boat, assuming that the overbearing rock does not obstruct the view? The index of refraction of water is n=1.33. (1) 11.4m (2) 8.8m (3) 7.5m (4) 14.1m (5) any distance PHY 2049 Fall 2013 – Acosta, Woodard Exam 3 solutions Light will not refract through the surface of the water if the angle it makes with respect to the normal to the surface is larger than the critical angle (in which case it is totally internally reflected). This angle is given by: n
1
⇒ θ crit = 48.75!
sin θ crit = 1 =
n2 1.33
w
tan θ crit = ⇒ w = 11.4m
d