PHYS 101 Light, Department of Physics
Prof. Ruiz/UNCA
EXAM 2 – Solutions
NO CALCULATOR
Name ___________________
March 26, 2013
Multiple Choice. Neatly mark your answer on the supplied scan card with a No. 2 pencil.
Q1. Additive mixtures of three colors are made by projecting colors
onto a white screen. The color in the “?” region is (A)black (B)cyan
(C)magenta (D)yellow (E)white. The “? is in the region where G
and R overlap. An additive mixture of green and red is yellow.
Q2. The hue angle in degrees for the color represented by the “?” at
the left is (A)0° (B)60° (C)120° (D)180° (E)240°. The hue angle
for yellow is 60°. Red is 0° and as you go counterclockwise you
get Y (60°), G (120°), C (180°), B (240°), and M (300°). Can you
give an easy way to remember these using complementary
colors?
Q3. Which of following could be the RGB value for the color designated by “?” in the figure? (A)00FF00
(B)00FFFF (C)FF0000 (D) FF00FF (E)FFFF00. The “?” color has to be a yellow. Since the G and R
circles overlap here, you go with the answer that has much with R and G. So you can get the answer
without even knowing about yellow. We rule out (A) since RGB = 00FF00 is a green only, (B) is
blue-green, i.e., cyan, (C) is red, (D) is R and B, i.e., magenta. The answer is (E) where RGB =
FFFF00, i.e., R = FF and G = FF, which only has R and G.
Q4. You set up the additive color-mixing experiment illustrated in the above figure using three overhead
projectors. You find that your red beam is not as strong as it should be. Therefore, the hue angle you gave
in Q2 will (A)decrease (B)remain the same (C)increase. If red is not as strong, this means the green
will be more dominant and we would have a greenish yellow rather than yellow. This means a shift
on the color wheel towards the green. Our hue angle will increase from 60° somewhat in the
direction towards 120° (green).
Q5. What is the complementary color for the “?” color in the above figure? (A)blue (B)cyan (C)green
(D)magenta (E)red. Method 1. The complementary color for yellow is blue. Method 2. Note that “?”
is opposite blue in the figure. When the blue circle overlaps the “?” region you get white. Since blue
and yellow produce white, they are complementary colors.
Q6. For your complementary colors in Q5 subtract the hue angle of one of them from that of the other
such that your result is a positive angle. Your result is (A)0° (B)45° (C)90° (D)150° (E)180°. Method 1.
Complementary colors are 180° apart from each other so the answer is 180°. Method 2. The hue
angle for blue is 240° and that for yellow is 60°. Therefore, 240° − 60° = 180°.
Q7. You replace the RGB beams in the above diagram with their respective complementary colors. Still
using rules for additive color mixing, which of the following could be the RGB value for the color
designated by “?” in your new modified version of the above figure? (A)AAAAFF (B)AAFFAA
(C)FFAAAA (D)AAFFFF (E)FFAAFF This means we are overlapping a magenta (complement for
green) and cyan (complement for red). Magenta is B + R and cyan is B + G. So we have one of each
with a some extra B. This is a bluish white. Go for the one that has mostly blue but also some green
and red. Since FF is max we have to pick AAAAFF where R = AA, G = AA, and B = FF.
Q8. You replace the RGB beams in the above diagram with their respective complementary colors. But
now, instead of addition, you use the rules for subtractive color mixing. Which of the following could be
the RGB value for the color designated by “?” in your modified version? (A)0000FF (B)00FF00
(C)FF0000 (D) FF00FF (E)FFFF00 If we subtractively mix magenta and cyan the result is blue.
Q9. The red component of the 6-bit color scheme (EGA) varies from 0 to a decimal maximum of (A)3
(B)15 (C)31 (D)63 (E)255. Method 1. Six-bit color is EGA, where each component can be 0 (off), 1
(low), 2 (medium), or 3 (high). Method 2. Six-bit color means two bits for each component. So the
red component can be 00 (off), 01 (low), 10 (medium), or 11 (high), where we are thinking in binary
instead of decimal. Method 3. Six-bit color is EGA and I remember EGA has 64 colors since 4 x 4 x
4 = 64. Therefore, each component has 4 levels. Counting zero as a level, the max is 3 for each
primary component.
Q10. The artist known for 20th-century photo-realism is (A)Constable (B)Flack (C)Manet (D)Monet
(E)Seurat.
Q11. The 19th century artist known for pointillism (color mixing by addition) is (A)Constable (B)Flack
(C)Manet (D)Monet (E)Seurat.
Q12. The branch of optics that deals with light traveling in straight lines rather than wave effects is called
(A)geometrical (B)physical (C)quantum optics. Think geometry – drawing rays as straight lines.
Q13. A point light source illuminates a 2-cm ball located 4 cm from the source. What is the diameter of
the shadow on a screen 4 cm behind the ball? (A)2 cm (B)4 cm (C)8 cm (D)10 cm (E)12 cm Method 1.
Sketch it. Method 2. It will be double due to similar triangles (do a sketch where you draw a vertical
diameter on the ball and on the shadow region).
Q14. Consider a solar eclipse. If the Moon were moved to be farther from the Earth, the diameter of the
umbra reaching the Earth would (A)decrease (B)remain the same (C)increase. If the moon were too far
from the Earth, the shadow would not even reach the Earth.
Q15. During totality of a total lunar eclipse the Moon can appear with a
(A)bluish (B)greenish
(C)reddish tint
due to a combination of light scattering in the Earth’s atmosphere and refraction.
Method 1. I remember seeing photos in class and in the e-text. Method 2. I remember discussion
about a reference to a blood-red moon in the Bible and astronomers attempting to date the
crucifixion using lunar eclipse dates. Method 3. When light goes through the atmosphere, the blue
scatters. When light goes through much atmosphere like at sunset the light that gets through and
hits clouds is red since so much blue is scattered away as well as some green. Light hitting the
atmosphere at an angle refracts towards the normal. So on its way to the moon light is refracted
towards the moon. The moon can appear reddish or orange due to the missing blue.
Q16. In constructing a pinhole camera, you make many aperture sizes starting at 0.5 cm and decreasing
the diameter in a series of experiments. The sharpness of the image (A)always improves when the your
pinhole aperture gets smaller and smaller (B)improves up to a point when the pinhole aperture gets
smaller and smaller (C)always deteriorates as the aperture gets smaller from your starting diameter
(D)deteriorates as the aperture gets smaller the begins to improve (E)improves as the aperture gets
smaller, then deteriorates, then finally improves. The image improves in sharpness as the pinhole gets
smaller from starting with a large opening. However, if you get too small, diffraction kicks in and
the light spreads out making the image fuzzy.
Q17. An image is formed in a pinhole camera. If you increase the distance between the pinhole and back
of the pinhole camera where the image is located, the size of the image (A)decreases (B)remains the same
(C)increases. Sketch the two cases.
Questions Q18 through Q20 refer to the figure at the left. There are
two mirrors at a right angle with a sketch of an individual (D)
standing in front of the mirrors.
Q18. Which images are virtual? (A)Only A (B)Only B (C)Only C
(D)Only A and C (E)Only A, B, and C All the images, i.e., Image
A, Image B, and Image C are virtual. You cannot form a real
image with a plane (i.e., flat) mirror.
Q19. For which images does D see her face as others see her. (A)Only
A (B)Only B (C)Only C
(D)Only A and C (E)Only A,
B, and C Method 1. Let the real person’s (D) right ear be the
little bump on the circle. If she touches her right ear, the
image B likewise must touch her right ear. In cases A and C it
is the left ear that is touched in a scenario like we saw in the
movie “Duck Soup.” Method 2. One reflection, the usual case,
leads to the “left-right reversal.” Therefore, for a double
reflection, i.e., two reflections (as is the case for B) you get two
reversals and these put things back. It is similar to the case
where two negatives make a positive. See right figure.
Q20. For which images does light reflect twice? (A)Only A
(B)Only B (C)Only C (D)Only A and C (E)Only A, B, and C See the rays in the above diagram.
Q21. The angle between the two mirrors for the kaleidoscope at the
right is (A)30° (B)36° (C)45° (D)60° (E)90°. These can be tricky.
There are 10 items in the pattern. But suppose you count 5 quickly.
Then your answer is 360°/5 = 72°. It is good that this answer is not
there since this is the wrong answer. The correct answer is 360°/10
= 36°. To help you we did not include 72° in the list of choices.
Q22. The magic trick at the left is
called the “Box of Pain.” The lady’s
head can be made to vanish most
effectively by using (A)one plane
mirror perpendicular to each side wall (B)two mirrors forming a “V”
(C)a half-silvered mirror.
See the left photo for how
things appear when the Vmirrors are in place. This is an
application of the general “V”
principle as seen in the right
figure.
The observer sees the walls
instead of the face or what is
hidden behind the mirrors.
Q23. The surface of the pool ball in the left photo is (A)concave
(B)convex (C)elliptical (D)parabolic (E)planar. Convex surfaces bow
outward.
Q24. The images of the fluorescent lights and pool player, actor Paul
Newman (1925-2008), are located (A)above the black ball (B)below
the black ball (C)in front of the black ball (D)inside the black ball
(E)outside the black ball behind its rear surface. For a convex mirror
all images are located behind the mirror between the mirror and the point F. Clearly this is inside
the ball. The ball is defined by completing the circle of the mirror around the center point C.
The Escher at the right is another example. Note that all images are
smaller except for the parts of the hand touching the mirror.
Q25. The images for Q24 are (A)real (B)virtual. Method 1. When rays are extended with dotted lines
we have a virtual image. Method 2. All images located behind any type of mirror are virtual.
Method 3. All images formed with a mirror in standard position are virtual when the images are
upright and real when the images are inverted.
Q26. In the left image, your instructor’s son is shooting the
reflection of himself due to a (A)concave (B)convex (C)planar
mirror. You can’t get an inverted image with the other choices
when everything is oriented vertically.
Q27. The videographer, whose image is smaller than life size, must
be (A)beyond twice the focal length away (B)between twice the
focal length and the focal length away (C)at the focal point (D)less
than the focal length away. The image has to be in front of him to
take the video. With an image smaller and inverted, the image must be between the focal point F
and center of curvature C. This places the object (the videographer) beyond “C,” i.e., beyond twice
the focal length away. The other case of inverted image would be if the videographer were between
C and F. But then the inverted image would be larger and far from the mirror – way behind the
videographer – out in the backyard somewhere.
Q28. The image made by the mirror in the above photo is (A)real (B)virtual. All inverted images are
real when formed by a single optical element where both the optical element and object are vertical.
Q29. The hand holding the camera (in the above image) is the videographer’s (A)left hand (B)right hand.
Draw an object arrow and a smaller inverted image due to a concave mirror. Sketch another arrow
into your page at the base of your first object arrow. Convince yourself that the image arrow comes
out of the page. From this, you should be able to figure out the answer.
Questions Q30 through
Q32 refer to the photos
of a ball that swings
towards and away from
the concave mirror. In
the far left photo the ball
is at that moment
farthest from the mirror.
At that moment, you can
take the image size to be
about the same size of the ball. In the right photo the ball is closest to the mirror. You can assume that the
image in the right photo is real.
Q30. The image of the ball in the left photo must be (A)real (B)virtual. The only way to obtain a virtual
image equal in size compared to the object is for the ball to touch the mirror surface. The ball never
gets that close since it is given for the closest point that the image is still real.
Q31. The image of the ball in the left photo must be (A)inverted (B)upright. All real images for a
concave mirror are inverted.
Q32. When the red ball is not swinging, but rather at rest, the center of the ball is (A)at the focal point F
(B)at the center of curvature C (C)somewhere between C and F (D)somewhere between F and the
mirror (E)somewhere farther from the mirror than the center of curvature C. When the ball swings
farthest, it reaches C since the real image is the same size as the object. When it swings between C
and F you get the larger real images. So at rest it must be between C and F.
Q33. What is the diopter value for a lens with focal length f = 50 cm? (A)0.5D (B)2D (C)5D (D)20D
(E)50D You express f in meters first to arrive at f = 0.5 m. Then you divide this into 1 and obtain 2.
You can think of this with an analogy using money: $1.00/$0.50 = 2.
Q34. What is the focal length if you place two lenses together where each separately has a focal length of
10 cm? (A)5 cm (B)10 cm (C)20 cm (D)25 cm (E)100 cm Method 1. Combining two converging lenses
makes a stronger lens. This means you have to end up with a focal length less than 10 cm. There is
only one choice. Method 2. You first need to express the lenses in diopters. Each is 10D so that the
combined result is 20D. A 20D lens has a focal length of 1/20 = 0.05 m = 5 cm.
Q35. What is the focal length in centimeters if you place a 10-cm focal length lens together with a −10-cm
focal length lens? (A)0 (B)10 (C)20 (D)100 (E)infinity Method 1. A converging lens combined with a
diverging lens of equal but opposite strength produces a neutral effect similar to a pane glass
window. Such has an infinite focal length. Method 2. You are combining a 10D lens with a −10D
lens. We can add diopters. The total is zero. A zero-strength lens has an infinite focal length.
Q36. When parallel white light enters a converging lens, chromatic aberration causes the blue light to
focus (A)closer to the lens (B)at the same distance from the lens (C)farther from the lens when you
compare the blue focus to the point where the red light focuses. The rule is “blue bends more” or you
can say “red bends less.”
Questions 37 through 43 refer to the figure below where you find two lenses. The left lens is referred to as
the first lens and we sketch it black as well as its focal points. The right lens, referred to as the second
lens, is sketched in red to help you identify its focal points as distinct from those for the first lens. Note
that the location of the second lens is at the second “black” focal point for the first lens.
Sketch a ray leaving the top of the arrow and going through the first focal point of the first lens, i.e., the
focal point that is to the left of the first lens.
Q37. Continue sketching what happens when your ray reaches the first lens. After passing through the
first lens, your ray between the two lenses (A)slopes downward (B)is parallel to the optic axis (C)slopes
upward. See the ray going through F on the left side of the lens. This is Ray 3 – you go through the F
on the left side of the lens, hit the lens, and go out parallel to the axis.
Q38. Continue your ray as it passes through the second lens. Your final outgoing ray (A)is parallel to the
optic axis (B)crosses the optic axis between the second lens and the second red focal point (C)crosses the
optic axis at the second red focal point (D)crosses the optic axis beyond the second red focal point
(E)does not cross the optic axis after leaving the second lens. Your parallel ray (second part of Ray 3)
between the two lenses becomes a Ray 1 with respect to the second lens. Therefore, you refract at
the second lens to go through the right focal point of the second lens.
Sketch a ray leaving the top of the arrow parallel to the optic axis and heading toward the left lens.
Q39. Continue sketching what happens when your ray reaches the first lens. Your ray sketched between
the two lenses (A)slopes downward (B)is parallel to the optic axis (C)slopes upward. See the parallel
ray leaving the object. This is Ray 1. It refracts heading towards the right focal length of the first
lens (see the second black F).
Q40. Continue your ray as it passes through the second lens. Your final outgoing ray (A)is parallel to the
optic axis (B)crosses the optic axis between the second lens and the second red focal point (C)crosses the
optic axis at the second red focal point (D)crosses the optic axis beyond the second red focal point
(E)does not cross the optic axis after leaving the second lens. Your ray is now a Ray 2 with respect
to the second lens. Therefore it passes straight through without changing direction.
Q41. Compared to the object, the image formed by the two lenses is (A)inverted and smaller (B)inverted
and larger (C)the same size (D)upright and smaller (E)upright and larger.
Q42. The location of the image formed by the two lenses is (A)to the left of the first lens (B)between the
lenses (C)between the first black focal point and the first red focal point (D)between the lenses, but
between the red “F” at the midpoint between the lenses and the second lens (E)beyond the second lens.
Q43. If you move the object farther from the lens system, the final image formed (A)decreases in size
(B)stays the same size (C)increases in size. As the object is moved to an infinite distance to the left,
the image gets smaller and smaller approaching the right focal point of the first lens (the second
black F).
Q44. Spherical aberration for a converging lens means that the marginal rays focus (A)closer to the
lens (B)at the same distance from the lens (C)farther from the lens
when you compare the marginal
rays with those hitting the inner region of the lens. The marginal rays bend too much.
Q45. When light passes through a birefringent material like calcite you
find a (A)single refraction (B)double refraction (C)triple refraction. The
two basic polarization states refract differently.
Q46. Light reflected from glass is polarized when the angle of incidence is
(A)nearly normal (B)18° (C)56°. This is the Brewster angle.
Q47. When photographing a rainbow the sun needs to be (A)to your left
(B)to your right (C)in front of you (D)behind you. Why? See e-text.
Q48. For which atmospheric effect listed does light actually penetrate
through ice crystals (A)sub sun (B)sun dog (C)sun pillar (D)rainbow.
Why? See e-text.
Q49. Which color is on top in a secondary rainbow? (A)blue (B)green (C)orange (D)red (E)yellow
Opposite to the one shown in this primary rainbow (see photo). Why? See e-text.
Q50. The mural shown of the archangel below where you also see the artist William Cochran is an
example of (A)anamorphic art, a form of art that dates back centuries (B)fauvism, an art movement
that began c. 1900 (C)op art that dates to the 1960s (D)photo-realism, a movement that began in the late
1960s (E)pointillism, a technique popularized during the latter 19th century.