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This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.
001 10.0 points Consider a prism with the shape shown in the diagram. Its index of refraction is labeled by np, and it is submerged in a special liquid which has an index of refraction nℓ, where the difference np − nℓ = 0.53. The light ray is perpendicularly incident from the liquid into the prism as shown in the diagram. Notice that the incident angle at both points A and B is 45◦.
np
A
B
nl
45 °
45 ° Find the index of refraction of the prism for a critical angle of 45◦.
Correct answer: 1.80953.
Explanation:
Let : ∆n = 0. 53 θc = 45◦^.
np sin θc = (np − ∆n) sin 90◦
np
= np − ∆n
np
= ∆n
np =
2 ∆n √ 2 − 1
002 10.0 points A diver shines a light up to the surface of a flat glass-bottomed boat at an angle of 44◦ relative to the normal. If the indices of refraction of air, water, and glass are 1.0, 1.33, and 1.6 respectively, at what angle does the light leave the glass (relative to its normal)?
Correct answer: 67. 5024 ◦. Explanation:
Let : na = 1. 0 , nw = 1. 33 , and θ = 44◦^.
air
glass
θw water
θa
Applying Snell’s law,
n 1 sin θ 1 = n 2 sin θ 2
Since
na sin θa = ng sin θg = nw sin θw ,
we do not need to consider glass in the solu- tion, and
θa = sin−^1
nw sin θw na
= sin−^1
1 .33 sin 44◦
- 0
= 67. 5024 ◦^.
003 10.0 points A light wave passes from air into water. Which of the following is true?
- Its wavelength remains unchanged, but its frequency decreases.
- Its frequency remains unchanged, but its wavelength increases.
- Both its wavelength and its frequency remain unchanged.
- None of these.
- Its wavelength remains unchanged, but its frequency increases.
- Its frequency remains unchanged, but its wavelength decreases. correct
Explanation: The frequency f of an electromagnetic wave is determined by the source producing it and does not change when the medium changes. The speed of an electromagnetic wave in water is v =
c √ K Km
where K is the dielectric constant and Km is the relative permeability. Since K > 1 and
Km > 1, c > v and the wavelength λ =
v f evidently decreases. Alternate Explanation:
f =
c λa =
vw λw
since c > vw, λa > λw.
004 10.0 points A certain kind of glass has an index of re- fraction of 1.65 for blue light and an index of 1 .615 for red light. If a beam containing these two colors in the air is incident at an angle 42. 2 ◦^ on this glass, what is the angle between the two beams inside the glass?
Correct answer: 0. 554645 ◦. Explanation: θi = 42. 2 ◦^ for both. θref r is different for the red and blue beams, so we use Snell’s law twice:
nair sin θi = nglass sin θref r
θref r = sin−^1
nglass
sin θi
θref r,red = sin−^1
sin 42. 2 ◦
= 0.428961 rad
θref r,blue = sin−^1
sin 42. 2 ◦
= 0.419281 rad
∆θ = θref r,red − θref r,blue = 0. 554645 ◦
005 10.0 points A fixed inductance 1. 27 μH is used in series with a variable capacitor in the tuning section of a radio. What capacitance tunes the circuit into the signal from a station broadcasting at 5 .9 MHz?
Calculate the radiation pressure acting on the cylinder.
- 8.10576e-
- 1.92773e-
- 2.31175e-
- 3.97929e-
- 3.20558e-
- 4.56751e-
- 7.24278e-
- 2.45396e-
- 6.50008e-
- 5.53755e-
Correct answer: 6. 50008 × 10 −^7 N/m^2. Explanation:
Let : r = 2.9 cm = 0.029 m , ℓ = 1.38 m , c = 2. 99792 × 108 m/s , and P = 6. 50008 × 10 −^7 N/m^2.
The intensity of the radiation reaching the walls of the cylinder is
I = 〈S〉 =
P
2 π r ℓ
so the radiation pressure on the walls is
p =
I
c
P
2 π r ℓ c
=
- 50008 × 10 −^7 N/m^2 2 π (0.029 m) (1.38 m) (2. 99792 × 108 m/s) = 6. 50008 × 10 −^7 N/m^2.
008 10.0 points A transformer has input voltage and current of 13 V and 5 A respectively, and an output current of 0.8 A. If there are 1297 turns turns on the sec- ondary side of the transformer, how many turns are on the primary side?
Correct answer: 207.52 turns. Explanation:
Let : ns = 1297 turns , Ip = 5 A , and Is = 0.8 A.
Energy is conserved, so
Pp = Ps Ip Vp = Is Vs Vp Vs
Is Is For the transformer
V ∝ n np ns
Vp Vs
Is Ip
n = ns
Is Ip
= (1297 turns)
0 .8 A
5 A
= 207 .52 turns.
009 10.0 points At what distance from a 39 W electromag- netic wave point source (like a light bulb) is the amplitude of the electric field 28 V/m? μo c = 376.991 Ω.
Correct answer: 1.72763 m. Explanation:
Let : P = 39 W , Emax = 28 V/m , and μ 0 c = 376.991 Ω.
The wave intensity (power per unit area) is given by
I = Sav
=
E max^2 2 (μo c)
=
(28 V/m)^2 2 (376.991 Ω) = 1.03981 W/m^2.
The constant μo c (impedance of free space) turns out to be 120 π Ω. You should verify that (V/m)^2 Ω = W/m^2. With this we can find the intensity, which multiplied by the area of a circle of radius R must gives us the source’s power P , therefore
4 π R^2 I = P = 39 W.
Solving for the radius we get,
R =
P
4 π I
39 W
4 π (1.03981 W/m^2 ) = 1 .72763 m.
010 (part 1 of 2) 10.0 points The magnitude of the radius of a spherical mirror is |R| = a. An object is placed in front of this spherical mirror. It forms a virtual
image. The magnification is M =
The mirror is
- convergent
- divergent correct
Explanation: Since there is a reduced vertical image, it must be a convex (divergent) mirror.
011 (part 2 of 2) 10.0 points The object distance is
a
a
a
a
a
- 3 a
a correct
- a
- 2 a
a
Explanation: Basic Concepts:
1 p
q
f
R
m =
h′ h
q p Concave Mirror f > 0 ∞ > p > f f < q < ∞ 0 > m > −∞ f > p > 0 −∞ < q < 0 ∞ > m > 1 Convex Mirror 0 > f ∞ > p > 0 f < q < 0 0 < m < 1
Solution: Since there is a reduced virtual image, it must correspond to convex mirror, or R = −|R|. Also, virtual image implies p q
p |q|
|M |
Hence, 1 p
q
R
p
|M |
|R|
p
a
−
p
a
keywords:
014 10.0 points The intensity of sunlight under the clear sky is 1150 W/m^2. How much electromagnetic energy is con- tained per cubic meter near the Earth’s sur- face? The speed of light is 2. 99792 × 108 m/s.
- 3.93606e-
- 2.8453e-
- 2.37831e-
- 3.25559e-
- 2.70187e-
- 3.76927e-
- 3.83599e-
- 3.53578e-
- 2.42835e-
- 3.43571e-
Correct answer: 3. 83599 × 10 −^6 J/m^3.
Explanation:
Let : I = 1150 W/m^2 and c = 2. 99792 × 108 m/s.
If v is the velocity at which the energy is transferred (the velocity of the wave), then the relationship between the intensity I (en- ergy per unit time per unit area) and the energy density of a wave u is
I = u c
u =
I
c
1150 W/m^2
- 99792 × 108 m/s = 3. 83599 × 10 −^6 J/m^3.
015 10.0 points The emf E can drive the circuit below at any given frequency.
E C L
R
A B
D
At what frequency does the light bulb glow most brightly?
- the frequency ω =
LC
- the frequency ω =
LC
- both very low frequencies or very high frequencies correct
- very low frequencies
- very high frequencies Explanation: Since the brightness of the bulb is propor- tional to the power dissipated in it,
P = I rms^2 R,.
In order to obtain maximum brightness we should adjust the frequency so that the rms current Irms through the bulb is the largest. The capacitive reactance XC =
ω C
is in- versely proportional to ω, and the inductive reactance XL = ω L is proportional to ω, so
Z = R +
XC
XL
= R +
ω C −
ω L
= R +
ω L √ ω^2 C L − 1
IR =
E
R +
ω L √ ω^2 C L − 1
ω Ilight bulb ω 0
ω 0 =
L C
Since the current is a minimum at reso- nance, the maximum brightness will occur at
both very low frequencies and very high fre- quencies.
016 10.0 points The reflecting surfaces of two intersecting flat mirrors are at an angle of 55◦, as shown in the figure. A light ray strikes the horizontal mirror at an angle of 51◦^ with respect to the mirror’s surface.
55 ◦^51 ◦
φ
Figure is not drawn to scale.
Calculate the angle φ.
Correct answer: 70◦.
Explanation: Basic Concept:
θincident = θref lected
Solution:
θ 1
θ 2
φ θ
Figure is to scale.
The sum of the angles in a triangle is 180◦^. In the triangle on the left we have angles
θ ,
180 ◦^ − θ 1 2
, and
180 ◦^ − θ 2 2
, so
180 ◦^ = θ +
180 ◦^ − θ 1 2
180 ◦^ − θ 2 2
, or
θ 1 + θ 2 = 2 θ. (1)
In the triangle on the right we have angles
θ 1 , θ 2 , and φ.
180 ◦^ = θ 1 + θ 2 + φ , so
θ 1 + θ 2 = 180◦^ − φ. (2)
Combining Eq. 1 and 2, we have
φ = 180◦^ − 2 θ = 180◦^ − 2 (55◦) = 70 ◦^.
As a matter of interest, in the upper-half of the figure the angles (clockwise) in the triangles from left to right are
39 ◦^ , 39 ◦^ , and 102◦^ ; 78 ◦^ , 35 ◦^ , and 67◦^ ; 113 ◦^ , 16 ◦^ , and 51◦^ ; 129 ◦^ , 16 ◦^ , and 35◦^ ;
and in the lower-half of the figure the angles (counter-clockwise) in the triangles from left to right are
16 ◦^ , 16 ◦^ , and 148◦^ ; 32 ◦^ , 35 ◦^ , and 113◦^ ; 67 ◦^ , 39 ◦^ , and 74◦^ ; 106 ◦^ , 39 ◦^ , and 35◦^.
017 10.0 points
Consider an electromagnetic wave pattern as shown in the figure below.
time (force per unit area) transferred to the wall is
∆ p ∆t A
F
A
2 I
c
2 P
A c
, so
F =
2 P
c
2 (0.021 W)
- 99792 × 108 m/s = 1. 40097 × 10 −^7 N.
