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Unformatted text preview: Name Section I’roblem 1: Short and Muitipie Choice Questions (Total 55) 1.1 A long thin light bulb iiiurninates a vertical aperture. Witth pattern of light do you
see on a viewing screen behind the aperture? (5) Light m (a) _ (b) .m em, an.qu «Faust"Jauva 1.2 A lens produces a sharply—focused inverted image on a screen. What will you see on
the screen if the lens is removed? (3) a.. The image will be inverted and blurry.
b. The image will remain the same but much dimmer. @There will not be image at ali.
d. The image will be right—side—up and sharp.
e. The image will be rightsidewup and blurry. Screen Lens _ Image Name Section 1.3 Two plane mirrors form a right angle. How many images of the ball you can see
in the mirror. Mark their approximate location in the picture. (6) “(were 1.4 You are looking at the Sun near your horizon but you see it higher (apparent
Sun). You know that the thinner the air the faster the speed of light (or equivalently
the smaller the index of refraction). Can you explain why you see the Sun higher in
the sky than its real position? (Make a sketch showing the ray path). (10) “1:43;: W
‘/ wuc we?» go». ; 9mm: amen“ M‘
Tmuu'a'at— AWL. .— ‘Fﬁ'wﬂs'ﬁ‘ quclﬂa'
OP my? ’T’IMQ. Name Section 1.5 An astronomer is trying to observe two distant stars. When he looks through a
green fiiter the stars are barely resolved. Should he order a blue or a red filter to
improve the resolution of his observations. Explain your answer. (Assume that the
his telescope resolution is not affected by the atmosphere) (3) ﬁgure: Warm — ozoneme unmade“ Q "A l '6 >5 6wcw‘f‘bau 3 (9%)
W“ t) 15 were u. were “mm” 1.6 Coherent monochromatic plane waves impinge on two apertures separated by
distance d. An approximate formula for the path length difference between the two
rays is (3) 3. dcosB
4. LcosB 1.7 The light passing through this slit when seen on a screen far from the slit will
exhibit destructive interference when (3) 1. (w2)sinemm
@(w2)sine=m
3. (oz/2)sin9£?L 4. (on/2)sinﬂ=?L/8 Name Section 1.8 Coherent monochromatic plane waves impinge on two long narrow (width a)
apertures separated by a distance d. The resulting pattern on the screen far away is shown below. For thi arrangement The value d/a is about:
(a) 1/8 (b) 1A (c) 4(e) cannot be determined. Justify your answer (10) Name Section 1.9 A paraxial ray (3) a. moves in a parabolic path.
h. is a ray that has been reﬂected from parabolic mirror.
@ is a ray that moves nearly parallel to the optical axis.
. ss a ray that oves exactly parallel to the optical axis. wA Virtual image is (3) a. the cause of optical illusions.
a point ﬁom which rays appear to diverge.
c. an image that only seems to exist.
d. the image that is left'in space after you remove a Viewing screen. 11.1 The focal length of a converging lens is (3) a. the distance at which an image is formed. ‘0. the distance at which an object must be placed to fonn an image.
the distance at which parallel light rays are focused.
. the distance from the ﬁont surface to the back surface. 1.12 A red card is illuminated with red light. What color will appear? What if is
illuminated with blue light? (explain your answer) (3) u » mu» HEM. ; « can chose Rescues»
233m) "i R. gnaw: BLUE. a) a] tree MW? 1‘; (Lasaacvats aw. was Clea—b
l”? luth AQMR 9.15:. Name Problem 2 £30): Consider an RLC circuit such as shown below with the following parameters: $0210 V, R=i0£2 and oo=23tf2100 c/sec. terminals A and B with both switches open? {5) 2.1: What are the values of the
maximum current i, voitage VAB
between the terminals A and B
and phase of VAs relative to the
phase of the applied EMF, when
the switches 51 and 52 are closed
(inductance and capacitance
shorted). (5) 2.2: Is there a combination of L
and C that will produce the same
current and voltage vaiues as in
question 2.1 with both switches
open? (Explain) (10) 2.3 What is the maximum average
power that the circuit can deiiver
to a motor connected at the 2.4 What is the maximum average power that the circuit will deliver to the motor
connected to the terminais A and D with both switches open and the terminal D disconnected? (Explain) (10) 2.2
Iago/Z
2.1 Z£VR2+(XL_X¢:)2
hag/1221A ¢ﬂmi(M)
rams IR a 10 Volts R
¢ # 0 For XL 3 X a (resonance)
mLﬂl/wc
I£IA,VAB :IOVﬁmO LC H 1/602 a 104 H emyFamd 2.3 £545— : SOWattS 2 P ” SRMSIRMS “ 2.4 At resonance the values of VL and Vc are cancelled out because they are anti—phased.
As a result the impedance from a to D is R, similar to A to B. The maximum average
power will still be 50 Watts. NAME: Probiern 3 (401: A laser shines light of wavelength 500 nm headon toward a diffraction grating containing 100
vertical slits spread eveniy over a horizontal distance of 1.00 miilirneter. A gray wali is 4.00
meters behind the diffraction grating from the laser. Let P denote the point on the wall where the
laser beam wouid have hit if the diffraction grating weren’t there. (2!) Your roommate says, “The reason there’s a bright spot at point P is because the light that
doesn’t undergo any diffraction or interference reaches P.” Do you agree? Disagree?
Explain. A bright spot appears at P not for the reason your roommate says, but rather, because 0?) constructive interference occurs there. Point P is approximately the same distance from
aii the slits; so crests that simultaneously pass through all the slits “meet up” at P. More
formally, at P, the path length difference for two adjacent slits is AI. = 0, which means
that point satisﬁes the constructive interference condition, AL u not for integer n, namely
11 = 0. You can reach the same conclusion starting with dsin '6: HA, since 1? is where 6 =
0. How far from point P is the center of the next nearest bright spot on the wall? Show your
reasoning. Start with the constructive interference condition dsin 6 "w" ml for integer n. Point P is where n w 0. So we need to ﬁnd the distance to the n m l bright spot (interference maximum). I’ll
ﬁnd that distance, x, by calculating the angle 9 and then using it to find x. Since 100 slits are spread out over 1.0 x 104’ meters, the distance between adjacent slits is d = (1.0 x i0”3
meters)/100 = 1.0 x 10""5 meters. The wavelength is 2. 3 500 nm 2 5.0 x 10’7 meters. So, for n = i, 6‘ works out to be quite smali, which means we can approximate sin 9 as 6:
6% sin 6: nA/d ﬂ (1)(5.0 x710"7 meters)/(l.0 x 10"5 meters) = 0.050 radians. For that smail angle, the iinear distance along the wall approximately equals the distance
along the corresponding arc length: x m R9 “—"* (4.00 meters)(0.050) m 0.20 meters. (c) All the slits are now covered except 2 neighboring slits near the center. Describe in what
ways the light pattern on the wall does and does not change, cementing speciﬁcally on
(i) its overall brightness, (ii) the spacing between the bright spots, and (iii) the width of
those bright spots. (i) The spots are much less bright, because only 2 slits—worth instead of 100 slits~worth of total light passes through the grating. ‘23“ a g as also *‘ a g d; c“ v: S? ‘5» o . (ii) The spacing stays the same, because hav1ng 100 (or 500 or 1000) slits reinforces the regions (iii) ((1) of constructive interference produced by just 2 slits instead of making a fundamentally
different interference pattern. The bright spots get wider (less crisp) because more slits contributing to destructive
interference between the bright spots results in wider regions of nearlyperfect destructive
interference and hence thinner regions of constructive interference. The slits are all uncovered again. But now the laser light is replaced with an equally thin
beam of light coming from a regular household light bulb. Does the same pattern of
bright spots appear on the wall as appeared in parts (a) and (b)? Explain. An interference pattern no longer appears. The “white” light from a regular bulb consists of many different wavelengths all mixed together, with no ﬁxed phase differences. So, it’s
no longer the case that when a crest reaches slit 1, a crest simultaneously reaches slits 2
through 100. By drawing the usual “overlapping circles” diagrams, you can see that a
stable interference pattern forms in the twoslit (or 100~slit) scenario only if the waves
reaching the diffraction grating are coherent, consisting of a single wavelength and with a
fixed phase difference between the waves reaching adjacent slits. Name: 1’roblem 5: An object of height h is placed in front of a convex mirror. The object is placed at
distance of twice the focal length. You may assume that the height of the object is much smaller
than radius of curvature of the mirror. A diagram of the set up is shown below; the dot represents
the focal point, the dashed line in the center is the axis of symmetry and the two clashed arrows
represent rays traveling from the top of the obj cot—«one traveling toward the focal point and the
other parallel to the symmetry axis. The mirror causes an image to be formed. (30) 5.1 On the diagram below continue the trace of the two rays including their reﬂections from the
mirror to find the outgoing rays. Indicate where the outgoing rays meet forming an image (30) 5.2 Algebraically solve for the distance the image is from the mirror using the appropriate
formula. Express your answer in terms of f Is this position consistent with your ray tracing in 5.1? (in) i 1 1
m+mmm
s s' f
S=2f
L_i__l
2f f s‘ Name: 5.3 Indicate whether the following statements are true or false. (You do not need to justify your
answers). (19) 1. The image is virtual. ii. The image is inverted.
iii. The image is an the opposite side of the mirror frem abject.
iv. The magnitude of the magnification is less than 2. v. The magniﬁcation is negativs. Items ii, iv, and v are tfue Name Section Problem 4: (5 6) Consider a “black box” which contains an
arrangement of components which may include resistors,
capacitors and inductors, the details of which you do net
know. Two wires 1ead into the box an and are hooked up to an AC voltage source with 5 = 50 cos(wr) as in the figure to the right: The current going into the black box is
measured as a function of time and is found to be of the 56 mo form: 1o) 2 (Am) cos(a)t) + 302)) Sin(a)t)) e
where AUG), 3(a)) are measured functions, L0 is a constant with dimensions of inductance  sjwko) a. Show that the average power dissipated in the black box is P = . ‘ Hint: The 2L0
easiest way to do this does not require computing the phase shift. C x o) T
7?" “L gicm edge“: a
a: T '0 «0 x; £3 wt”
'1' w ‘5me "
.. 5: wt— *5”; » 7% \W Name Section b. Although you cannot look into the black box, from the form of the measured current
outside one can deduce whether or nor the there is a resistor in the box hooked into the circuit. Is there a resistor? How do you know? (Hint: think about the result of part a.) z "" 'Powén. T'E'io 1"?
9mm: “rue AVEMEZ. .
were. A BEQWTUQU mwewc'rvws. ate‘me ces‘eecmes 9‘9 V0". 9 Weave1%. «Pavia . I g ‘ M214 gab)
I WW 5? Mesa— cg HM“: '3‘ germ; my” & 1‘33) H .0 C. Find an expression for the phase shift between current and voltage. Reoaﬂ that if the
voltage is written in the form 6 s 50 oos(an) and the current as 10‘) 2: I 0 COS(CUI —— 425)
then the phase shift isgﬂ Express your answer in terms of AUG) and Ma).
wish to use trig identities such as sin(a + b) : sin(a)cos(b) + cos(a) sin(b) or
00302 + b) = c0302) oos(b) — sin(a) 3111(1)) . g {53 You may ...
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This note was uploaded on 12/28/2011 for the course PHYSICS 270 taught by Professor Drake during the Fall '08 term at Maryland.
 Fall '08
 drake

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