Ch 13 Light & Reflection
Updated 5/20/07
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13.1
Characteristics of Light
Practice A, Electromagnetic Waves, pg 449
1] 1.0x10-13 2] 3.4 m -
2.78 m 3] 85.1 m - 10.1 m, The
wavelengths are shorter than
those of the AM radio band.
4] 3.0x105 Hz 5] 5.4x1014
Hz 6] 2.4 x 1015 Hz
Section 13.1 Review Q, pg 450:
1a] microwave [b] radio waves & visible light [c] visible light
2] 3.96x10-7 m, near
ultraviolet
3] The speed of light is too great to measure over such a short distance. The
time of travel for the light in Galileo’s experiment was about 1x10-5
s.
4] 1/16 of the sun’s brightness on Earth’s surface.
Ch 13 Review Q, pg 476:
1a] radio waves [b] gamma rays
2] ultraviolet radiation
3] Its speed is accurately known. Measuring the time it takes light to travel a
distance allows the distance to be determined. D = St
However, for distances
of 100 km (62 mi) or less:
t = D/S = 105m/3x108m/s = 1/3,000 s
That is, the timing device has to be very accurate.
Alternatively, if the source’s brightness (I) is known, its apparent brightness
(E) can be measured and its distance calculated:
E = I/R2 ---> R = √I/E
4] The wave front at B would be an arc of a large circle. The rays would point
radially outward from A to B.
5] Apparent brightness (E) equals the actual brightness (I) divided by the
square of the distance (R2)between observer and source:
E = I/R2
6] 1999 + 2(95) = 2189
7] The speed of light: 3x108
m/s
8] The light from galaxies was emitted millions of years ago.
9] No; those stars may be closer and so appear brighter.
10] V = λf ---> λ = V = 3.0x108
m/s = 4.0x10-7 m
f 7.5x1014
/s
and 3.0x108 m/s =
3.0x10-7 m
1.0x1015 /s
11] 1x10-6 m = 1 thousandth of
a millimeter
12] 3.02 m
13] 9.1x10-3 m = 9.1 mm
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13.2 Flat Mirrors
Section 13.2 Review Q, pg 454:
1a] specular [b] diffuse [c] specular [d] diffuse
2] 75o; 75o
3] Be sure that the rays’ angles of incidence = angels of reflection.
4] The light must be reflected from one mirror to the other in order to form
images of images. Therefore, the mirrors must be exactly or very nearly
parallel to each other, with their mirrored surfaces facing each other.
5] It will seem twice as large, because the object distance (the distance
between the mirror and any point in the room, including the back wall) equals
the image distance.
6] An object facing the mirror produces an image that faces the object, and the
front of the object corresponds to the back of the image. It appears as if the
image is an object located behind the mirror that is left-right reversed.
Ch 13 Review Q 14-22, pg 476-477:
14a] diffusely [b] specularly [c] specularly [d] diffusely
[e] specularly
15] You would point the mirror so its normal would bisect the angle between the
sun and aircraft.
16] 2 m behind; Magnification = 1
17] The gas molecules in air do not reflect light. The diameter of an oxygen
molecule is 370x10-12 m while
the wavelength of light is about 370 x 10-7m.
Thus you would need 370x10-7
m/370x10-12 m = 105
or 100,000 oxygen molecules stuck together (liquid O2) to reflect
light.
18] Reflection is diffuse if λ is smaller than surface irregularities of
reflector.
19] No; your diagram should show that the ray from feet reflected at bottom of
mirror goes to the top of the head above the eyes.
20] θ2’ = 55o; Ray reflected from the second mirror is
always parallel to the incoming ray.
21] 1.2 m/s; the image moves toward the mirror’s surface.
22] Images serve as objects for more images. Each reflection doubles the
apparent distance from “object” to mirror.
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13.3 Curved Mirrors
Practice B, Imaging with Concave Mirrors, pg 462
1] p = 10.0 cm: no image (infinite q); p = 5.00 cm: q = -10.0 cm,
M = 2.00; virtual, upright image.
2] q = 53 cm; M = -0.57, real, inverted image.
3] R = 100 cm; M = 2.00; virtual image
4] f = 6.00 cm; M = -1.20; q = 7.71 cm; M = -0.286; real image
Practice C, Imaging with Concave Mirrors, pg 466
1] p = 46.0 cm; M = 0.500; virtual, upright image; h = 3.40 cm
2] M = 0.04; p = 6 m; h = 2 m; virtual, upright image.
3] p = 45 cm; h = 17 cm; M = 0.41; virtual, upright image.
4] q = -0.25 m; M = 0.081; virtual, upright image.
5] q = - 1.31 cm; M = 0.125; virtual, upright image
6] q = -20 cm; M = 0.41; virtual, upright image
Section 13.3 Review, pg 468:
1] q = -0.45 cm; m = 0.41; virtual, upright image
2] Concave mirror; p = 0.14 m
3a] real [b] virtual [c] virtual
4] The rays reflected by a parabolic mirror all focus at one point, whereas the
rays reflected by a concave spherical mirror reflect along a line that includes
the mirror’s focal point.
5] 132.5 m, which is the telescope’s focal length
Ch 13 Review Q 23-36, pg 477-478:
23] Concave
24] Real, inverted image, in front
25] No, rays always diverge from a convex mirror.
26] no, h’ < h for convex mirrors
27] no spherical aberration
28] through the focal point
29] A real image appears at the former object position.
30] No; no; image is upright and virtual when p < f
31] The light is spread out more in the larger image.
32] Try to project image on paper.
33] Produce rays parallel to an far from the principal axis.
All rays focus at F for a parabolic mirror.
34a] M = -0.384; real, inverted
b] M = -1.00; real, inverted
c] M = +1.67; virtual, upright
35] q = 26 cm, ; real, inverted; M = -2.0
36] p = 52.9 cm; h = 5.69 cm; M = 0.299; virtual, upright
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13.4
Color & Polarization
Section 13.4 Review, pg 474:
1] blue
2] magenta; red
3] yellow and cyan (more yellow than cyan);
green and red (more green than red)
4] If you turn the polarizer and very little light is transmitted, then the
reflected light is polarized.
Ch 13 Review Q 37-45, pg 478:
37] Red, green, blue; they make white light.
38] Cyan, magenta, yellow; they make black pigment.
39] The polarized light from the first polarizer is blocked by the second
polarizer when the component of the light that is parallel to the second
polarizer’s transmission axis equals zero. The light must be perpendicular (90o)
to the second polarizer’s transmission axis.
40a] green pigment b] white light c] black pigment d] yellow light
e] cyan light
41a] magenta b] red c] blue d] black e] red
42] cyan; blue
43] Rotate the sunglasses while looking at the sky or sunlight reflecting off a
horizontal surface. If brightness changes, the glasses have polarizing lenses.
44] Light reflected from a horizontal surface like an auto hood is polarized
horizontally and is blocked by the lenses. Light reflected from tall narrow
surfaces like the tank will be vertically polarized, and almost all of it will
pass through the lenses.
45] Yes; light from the sky is polarized, but light from the clouds
is not polarized.
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Mixed Review
46] p = 410 cm; f = 32 cm; R = 64 cm; real inverted
image.
47] inverted; p = 6.1 cm; f= 2.6 cm; real
48] R = -31.0 cm
49] q2 = 6.7 cm; real; M1 = -0.57, M2 = -0.27; inverted
50] f = -13.7 cm; M = 0.0656; virtual, upright.
51] p = 11.3 cm
52] Concave; M = - 20.0; real; inverted
53]
54]
55] R = -25.0
56a] 15.0 cm
b] 59.9 cm
c] Mconvex = 2.00 Mconcave = 0.667
d] virtual
e] upright
57] Concave, R = 48.1 cm; M = 2.00; virtual
58]
59]
60]
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Standardized Test Prep, pg 482
1] B [2] H [3] C [4] H [5] B [6] F [7] A [8] G
[9] C [10] J 11] The blue fabric appears blue. The red fabric appears black.
12] 65o [13] 6x10-12
m = 6.0 pm
14] Polarized light will pass through the plastic when the transmission axis of
the plastic is parallel with the light’s plane of polarization. Rotating the
plastic 90o will prevent the polarized light from passing through the
plastic, so the plastic appears dark. If light is not linearly polarized,
rotating the plastic 90o will have no effect on the light’s
intensity.
15] 15.0 cm [16] 10.0 cm [17] -0.500 [18] -6.0 cm
19 real; inverted
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Notes: 13.2 Flat Mirrors
Obj: Distinguish between specular and diffuse reflection of
light.
1 DRAW A WAVEFRONT DIAGRAM
AND THE CORRESPONDING WAVE RAY FOR:
a] A SERIES OF STRAIGHT WAVES. (3 labels)
b] A SERIES OF CIRCULAR WAVES.
c] A STRAIGHT WAVE REFLECTED FROM A FLAT SURFACE. (8
labels)
http://sol.sci.uop.edu/~jfalward/physics17/chapter12/chapter12.html
d] STATE THE 2 LAWS OF REFLECTION.
Angle of Incidence = Angle of Reflection
The incident Ray, Reflected Ray and Normal lie in
the same plane.
2 Distinguish between specular and diffuse reflection of
light.
Specular reflection occurs at flat surfaces (mirrors & lakes), is without any
distortion, the image will stay in tact.
Diffuse reflection occurs at rough surfaces, is scattered and no image results.
Obj: Apply the law of reflection for flat mirrors.
3 a] Draw and label the diagram for a smart mirror.
See Lab Flat Mirror 1
b] Diagram a 6” arrow reflected in a flat mirror.
c] Find the location (Di), and size (Hi) of the image
d] Find the Magnification
e] State the type of image.
See Lab Flat Mirror 2
Obj: a] Describe the nature of images formed by flat
mirrors.
4 State the 5
Characteristics of a Flat Mirror.
a] Image size and object size are equal: Ho = Hi
b] Image is the same distance behind the mirror as the object is in
front: Do = Di
c] Image is virtual (behind the mirror)
d] Image is erect (points in the same direction as the
object)
e] Right & left are interchanged
(your right hand looks like your left)
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Obj: Calculate distances and focal lengths using the mirror
equation for concave and convex spherical mirrors.
Diagram Rules:
a] Color object and incident rays red.
b] Color image and reflected rays blue.
c] Incident rays start at the object (arrow) and end at the
mirror.
d] Incident rays run at all 360o. We are
interested in only two:
One that runs parallel to the axis and one that runs
inline with
the Focal Point.
e] Reflected rays start at the mirror and end off the page.
The incident ray parallel to the axis reflects inline with
the
focal point. The incident ray in line with the focal
point
reflects parallel to the axis.
f] Image head is formed where the reflected rays from the
head cross.
If they do not cross trace them back behind the mirror
until they do.
1 Derive the
2 mirror equations.
a] The flat mirror assumption: The mirror’s
radius is large compared to the size of
the object, i.e. the mirror is
assumed flat. Ho<<R
b] Draw & label the diagram and derive the two
mirror equations.
Diagram | Object
(on red margin 3 rows high) F |/Mirror
See Board
|<----------------4”----------------->+<----2”---->|/
c]
Law of Similar Triangles: If two triangles are similar then the ratios of their
corresponding sides are equal.
d] Apply Law of Similar Triangles
Red triangles: Ho = Hi
------> Hi = Di Eaa #1
Do Di Algebra Ho
Do
Blue triangles: Ho =
Hi -----> Do = Do - F ------> 1 = 1 + 1
Do - F F Eq #1
Di F Algebra F Do Di
2
Use the Mirror
Equation: 1 = 1 + 1 to solve for F, Do, & Di
F Do Di
1 = 1 + 1 Multiply both sides by FDoDi--->DoDi =
FDi + FDo
F Do Di Subtract FDi from both sides--->DoDi - FDi = FDo
Factor out Di --->Di(Do - F) = FDo
Divide both sides by Do - F --->Di = FDo
Do - F
Substitute this into the equation above.
Hi = Ho FDo = HoF
Do (Do - F) (Do - F)
Starting with --->DoDi = FDi + FDo
Subtract FDo from both sides--->DoDi - FDo = FDi
Factor out Do --->Do(Di - F) = FDi
Divide both sides by Di - F --->Do = FDi
Di - F
Starting with: DoDi = FDi + FDo
Factor out F ---> DoDi = F(Di + Do)
Divide both sides by Di + Do --->F = DoDi
Di + Do
Equation Summary
Hi = Di & 1 = 1 +
1 -------> Di = FDo Do
= FDi F = DoDi
Ho Do F
Do Di Algebra Do - F
Di - F Di + Do
Practice B
1 Find the image distance & magnification of the mirror
in the sample problem
(F = 10 cm) when the object distances are 10.0 cm and 5.00
cm
Di = FDo = 10x10 = 100 = Infinity ==>No image
Do - F 10-10 0
Di = FDo = 10x5 = 50 = -10 cm; Negative Di===>Virtual,
upright
Do - F 5-10 -5 image.
M = Hi = |Di| = 10 = 2
Ho |Do| 5
2 A concave shaving mirror has a focal length of 33 cm.
a] Calculate the image position of a cologne bottle placed at 93 cm.
Di = FDo = 33x93 = 3069 = 51 cm==>Since Di > F the
image is real
Do - F 93 - 33 60 and inverted
b] Calculate the magnification of the image.
M = Hi = |Di| = 51 = 0.55 ==> Image is reduced by about
half
Ho |Do| 93
3 A concave makeup mirror is designed so that a person
25.0 cm in front of it sees an upright image at a distance of 50.0 cm behind the
mirror. What is the radius of curvature of the mirror.
Given: Do = 25.0 cm, Di = -50.0 cm
Find: R = 2F
F = DoDi = 25(-50) = -1250 = 50 cm
Do + Di 25 + (-50) -25
R = 2F =2(50) = 100 cm
Person at 25 cm is between the focal point and Mirror==>image is upright &
virtual
b] Find the magnification:
M = Hi = |Di| = 50 = 2.0 ==> Image is enlarged
Ho |Do| 25
4 A pen placed 11.0 cm from a concave spherical mirror
produces a real image 13.2 cm from the mirror.
a] What is the focal length of the mirror?
Given: Do = 11.0 cm, Di = 13.2 cm
Find: F = ?
Estimated
F = DoDi = 11.0(13.2) ~ 10(12) = 120
= 20 = 5 cm
Do + Di 11.0 + 13.2 11 + 13 24 4
Calculated
F = DoDi = 11.0(13.2) = 145.2 = 6.00 cm
Do + Di 11.0 + (13.2) 24.2
b] If the pen is placed 27.0 cm for the mirror, what is the new position of the
image?
Given: Do = 27.0 cm, F = 6.00 cm
Find: Di = ?
Estimated
Di = FDo = 6.00(27) = 6(30) = 3x3 = 9.00
cm
Do - F 27.0 - 6 20
Calculated
Di = FDo = 6.00(27) = 162 = 7.71
cm
Do - F 27.0 - 6 21
c] Find the magnification:
M = Hi = |Di| = 13.2 = 1.2 ==> Image is enlarged by 20%
Ho |Do| 11
Obj: Draw ray diagrams to find the image distance and
magnification
for concave and convex spherical mirrors.
See Lab
Obj: Distinguish between real and virtual images.
Real images are in front of the mirror and can be projected onto a
screen.
Virtual images are in back of the mirror and cannot be projected onto a
screen.
Obj: Describe how parabolic mirrors differ from spherical
mirrors.
Rays that reflect at a spherical mirror’s surface far from the axis
intersect at
slightly different points on the axis and not at the focal point. This
produces a
blurred image called spherical aberration. Mirrors shaped
as a parabola do not have
this problem.
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