Version 001 – Optics Review Quest – tubman – (20131B)
This print-out should have 19 questions.
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before answering.
Vertical Flat Mirror
001 10.0 points
Hint: A ray diagram would be helpful.
Determine the minimum height of a vertical
flat mirror in which a person 62 in. in height
can see his or her full image.
Your answer must be within ± 1.0%
Correct answer: 31 in..
Explanation:
In the figure, the mirror is labeled AB.
A ray from the woman’s foot F strikes the
bottom of the mirror at B, with an angle
equal to θ and proceeds to the woman’s eye.
D
T
A
E
L
M
F
θ
θ
B
Christmas Tree Ornament 02
002 (part 1 of 2) 10.0 points
An object is 21.4 cm from the surface of a
reflective spherical Christmas-tree ornament
7.5 cm in diameter.
What is the position of the image?
Your answer must be within ± 1.0%
Correct answer: −1.72395 cm.
Explanation:
1 1
1
2
h′
q
+ = =
m=
=−
p q
f
R
h
p
Convex Mirror 0 > f
∞ >p> 0 f <q< 0
1
EM = M F = EF
2
which is also the distance BC. Similarly, a ray
from the top of the woman’s head T strikes
the top of the mirror at A and proceeds to
her eye. The same line of reasoning as above
leads to the conclusion that
1
DA = T E
2
1
(T E +
2
EF ), which is one half the woman’s height.
Note that the mirror’s bottom edge must be
1
exactly EF from the floor for a full-height
2
image to be possible. Note also that the
conclusions reached here are valid regardless
of how far she stands from the mirror.
Thus the length AB of the mirror is
0 <m< 1
Let : D = 7.5 cm and
p = 21.4 cm .
p is positive since it is in front of the mirror
and R is negative since it is behind the mirror. A spherical Christmas-tree ornament is a
convex mirror, so
C
The two right triangles EBM and F BM
are identical, since they share the common
side M B and angle θ. Therefore
1
2
1 1
+ =−
.
p q
|R|
We are given the object distance, p, and can
find the radius of curvature from the diameter,
|D|
|R| =
. Inserting these values into the
2
mirror equation and solving for q, we find
q=−
=−
1
1
4
+
D p
1
4
1
+
(7.5 cm) (21.4 cm)
= −1.72395 cm .
003 (part 2 of 2) 10.0 points
What is the magnification of the image?
Your answer must be within ± 1.0%
Correct answer: 0.0805585.
Explanation:
Version 001 – Optics Review Quest – tubman – (20131B)
The magnification is given by
q
−1.72395 cm
M =− =−
= 0.0805585 .
p
21.4 cm
Object Moving Closer 02
004 (part 1 of 4) 10.0 points
The goal of this problem is to describe the image of an object as it moves from far away
towards a convex mirror of radius R.
′
7. s = f
s−f
s
2
Explanation:
The formula for a convex mirror is the same
as for a concave mirror,
1
1
1
+ =
′
s
s
f
Rearranging, we get
For a convex mirror, the focal length is
1
s−f
=
.
′
s
fs
1. f = R/2 and is negative. correct
Thus,
2. f = 2R and is negative.
′
s =f
s
s−f
3. f = R and is positive.
4. f = R/2 and is positive.
5. f = 2R and is positive.
6. f = R and is negative.
Explanation:
R
The focal length is f = . For a convex
2
mirror, R < 0, so f is also negative.
006 (part 3 of 4) 10.0 points
What is the formula for m, the magnification,
and how does it relate to the orientation of
the image?
1. m = −s′ /s, if m < 0 the image is inverted
correct
2. m = −s′ /s, if m > 0 the image is inverted
3. m = s′ /s, if m < 0 the image is inverted
005 (part 2 of 4) 10.0 points
The distance s′ of the image to the mirror is a
function of the distance s of the object to the
mirror and the focal length f of the mirror.
What is this function?
s+f
′
1. s = f
s
2. None of these.
s
′
3. s = f
f −s
s
′
4. s = f
s+f
s
′
correct
5. s = f
s−f
f −s
′
6. s = f
s
4. m = s′ /s, if m > 0 the image is inverted
Explanation:
s′
m = − , just as in the concave mirror case.
s
If m < 0, the image is inverted.
007 (part 4 of 4) 10.0 points
When s > 0, what is s′ , and what is the image
like?
1. s′ > 0, image is smaller than object, and
inverted
2. s′ < 0, image is bigger than object, and
Version 001 – Optics Review Quest – tubman – (20131B)
3
4. Unable to determine.
erect
5. converging lens
′
3. s > 0, image is smaller than object, and
erect
6. converging mirror correct
Explanation:
4. s′ > 0, image is bigger than object, and
erect
5. s′ > 0, image is bigger than object, and
inverted
The parallel light rays reflect and pass
through a single point, which must be the
focal point of a converging (concave) mirror.
6. s′ < 0, image is smaller than object, and
inverted
Spherical Mirror B 01
009 10.0 points
A concave spherical mirror forms a real image
2.01 times the size of the object. The object
distance is 27.6 cm .
7. s′ < 0, image is smaller than object, and
erect
correct
8. s′ < 0, image is bigger than object, and
inverted
h
q
R
Explanation:
From the graph, when s > 0, s′ < 0. For
all values of s > 0, |s′ | < s, so the image is
s′
smaller than the object. As m = − > 0,
s
the image is erect.
Ray Diagrams
008 10.0 points
Consider the light rays depicted in the figure.
What type of reflecting or refracting surface
is depicted?
p
f
h′
Scale: 10 cm =
Find the magnitude of the radius of curvature of the mirror.
Your answer must be within ± 1.0%
Correct answer: 36.8611 cm.
Explanation:
1 1
1
2
h′
q
+ = =
M=
=−
p q
f
R
h
p
Concave Mirror
f >0
∞ >p> f
f <q< ∞
0 > M > −∞
1. plane mirror
2. diverging lens
3. diverging mirror
Note: The radius of curvature for a concave
mirror is positive.
q
M = − = −2.01 .
p
(1)
Version 001 – Optics Review Quest – tubman – (20131B)
Solving for q, we have
q = −M p
= −(−2.01) (27.6 cm)
= 55.476 cm .
(2)
Basic Concepts: Snell’s law, Total internal reflection.
Solution: Let x1 be the horizontal distance
from the laser to where the laser beam strikes
the water and x2 the horizontal distance from
that point to Robin (see the following figure).
Substituting these values into the mirror
equation
=
2
1 1
+
p q
(3)
2
1
1
+
(27.6 cm) (55.476 cm)
y2
Then we have
y1
tan θ1
(1.45 m)
=
tan(27.9◦ )
= 2.73857 m .
x1 =
From Snell’s law,
nair sin(90◦ − θ1 ) = nwater sin θ2 .
So
sin(90◦ − 27.9◦ )
(1.333)
= 0.66299
θ2 = arcsin(0.66299)
= 41.5283◦ .
sin θ2 =
Hence,
water
y2
R
x
B
How far (horizontal distance) is Robin from
the edge of the pool? (Fear not, Batfans. The
“R” is made of laser-reflective material.)
Your answer must be within ± 1.0%
Correct answer: 6.1393 m.
Explanation:
θ2
x2
θ1
Mirrored
Surface
Batplastic
surface
90ο−θ1 y1
R
Batman and Robin a
010 10.0 points
Batman and Robin are attempting to escape
that dastardly villain, the Joker, by hiding
in a large pool of water (refractive index
nwater = 1.333). The Joker stands gloating at
the edge of the pool. (His makeup is watersoluble.) He holds a powerful laser weapon
y1 = 1.45 m above the surface of the water
and fires at an angle of θ1 = 27.9◦ to the horizontal. He hits the Boy Wonder squarely on
the letter “R”, which is located y2 = 3.84 m
below the surface of the water.
y1
x1
air
water
= 36.8611 cm .
J
θ1
J
R = 2f
=
4
x2 = y2 tan θ2
= (3.84 m) tan 41.5283◦
= 3.40073 m .
Therefore, the distance of Robin from the
edge of the pool is
x1 + x2 = 2.73857 m + 3.40073 m
= 6.1393 m .
Focal Length of a Thin Lens
011 (part 1 of 2) 10.0 points
Version 001 – Optics Review Quest – tubman – (20131B)
An object located 33.3 cm in front of a lens
forms an image on a screen 8.21 cm behind
the lens.
Find the focal length of the lens.
Your answer must be within ± 1.0%
Correct answer: 6.5862 cm.
Two situations lead to virtual images.
For a divergent lens,
f
Explanation:
Let :
p = 33.3 cm and
q = 8.21 cm .
1
1 1
q+p
= + =
f
p q
pq
pq
(33.3 cm) (8.21 cm)
f=
=
p+q
33.3 cm + 8.21 cm
012 (part 2 of 2) 10.0 points
What is the magnification of the object?
Your answer must be within ± 1.0%
Correct answer: −0.246547.
Explanation:
Magnification is
q
8.21 cm
M =− =−
= −0.246547 .
p
33.3 cm
Virtual Image and Lenses
013 (part 1 of 2) 10.0 points
The light rays from an upright object when
passing through a lens from left to right lead
to a virtual image. The absolute value of the
magnification of this image is greater than
one.
Select the correct statement.
1. The lens can only be a divergent lens.
2. The lens can either be a convergent or a
divergent lens.
3. The lens can only be a convergent lens.
correct
f
q
p
for any p > 0, the image is always located
on the same side of the lens (virtual) with
the image closer to the lens than the object
(|q| < p), with the size of the image always
reduced.
On the other hand for a convergent lens,
= 6.5862 cm .
Explanation:
5
f
f
p
q
when an object is placed within the focal
point (0 < p < f ), the corresponding image
is located on the same side of the lens (virtual) and outside of the focal point (|q| > f ),
leading to a larger image.
014 (part 2 of 2) 10.0 points
What is the nature of this virtual image?
1. Upright and located to the left of the lens
correct
2. Upright and located to the right of the
lens
3. Inverted and located to the left of the
lens
4. Inverted and located to the right of the
lens
Explanation:
Lens Convergent Diagram
015 10.0 points
Version 001 – Optics Review Quest – tubman – (20131B)
6
A virtual object is located to the right of
a convergent lens. The object’s distance and
image’s distance from the lens and the lens’
focal length are shown in the figures.
3.
f
f
f
f
17
f0
28
17
f
11
Which diagram correctly shows the image?
The convergent lens in this problem is a part
of a lens system so the object in this problem
may be either real or virtual. Construct a ray
diagram.
17
f
11
0
4.
f
f
0
Explanation:
1.
f
f
q f p
f
17
f0
28
17
f
11
0
2.
f
f
0
correct
17 17
f
f
28 11
17 17
f
f
28 11
17 −17
f
f
28 11
1 1
1
+ =
p q
f
1 1
−11 1
1
=− + =−
+
q
p f
17 f
f
−(−11) + 17
28
=
=
17 f
17 f
17
f.
q=
28
The magnification of this lens is
Version 001 – Optics Review Quest – tubman – (20131B)
17
f
q1
−11
11
28
m=− =−
=−
=
.
17
p1
28
28
f
−11
Concave Mirror Image 02
016 (part 1 of 4) 10.0 points
A certain concave spherical mirror has a focal
length of 10.5 cm.
Find the location of the image for an object
distance of 26.6 cm.
Your answer must be within ± 1.0%
Correct answer: 17.3478 cm.
7
means that the image is inverted. Finally,
because q1 = 17.3478 cm is positive, the image
is located on the front side of the mirror and
is real.
018 (part 3 of 4) 10.0 points
Find the location of the image for an object
distance of 3.23 cm.
Your answer must be within ± 1.0%
Correct answer: −4.66506 cm.
Explanation:
Let :
p2 = 3.23 cm .
Equation (1) implies
Explanation:
p2 f
p2 − f
(3.23 cm) (10.5 cm)
=
3.23 cm − 10.5 cm
q2 =
1 1
1
2
h′
q
+ = =
m=
=−
p q
f
R
h
p
Concave Mirror
f >0
∞ >p> f
f >p> 0
f < q < ∞ 0 > m > −∞
−∞ < q < 0 ∞ > m > 1
Let : p1 = 26.6 cm and
f = 10.5 cm .
1
1
1
= −
q1
f
p1
p1 f
q1 =
p1 − f
(26.6 cm) (10.5 cm)
=
26.6 cm − 10.5 cm
= 17.3478 cm .
017 (part 2 of 4) 10.0 points
What is the magnification for an object distance of 26.6 cm?
Your answer must be within ± 1.0%
Correct answer: −0.652174.
Explanation:
The magnification is given by
M1 = −
q1
17.3478 cm
=−
= −0.652174 .
p1
26.6 cm
This value of M1 means that the image is
smaller than the object. The negative sign
= −4.66506 cm ,
that is, in this case the image is virtual because it is located behind the mirror.
019 (part 4 of 4) 10.0 points
Calculate the magnification for an object distance of 3.23 cm.
Your answer must be within ± 1.0%
Correct answer: 1.44429.
Explanation:
The formula for the magnification yields
M2 = −
q2
−4.66506 cm
=−
= 1.44429 .
p2
3.23 cm
From this, we see that the image is 1.44429
times as large as the object and the positive
sign for M2 indicates that the image is upright. The negative value of q2 means that
the image is behind the mirror and is virtual.