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Homework 13
chapter 36: 17, 21, 33, 74
Problem 36.17
A spherical mirror is to be used to form, on a screen located 5 m from the object,
an image five times the size of an object. (a) Describe the type of mirror required.
(b) Where should the mirror be positioned relative to the object?
One can solve the problem formally from the mirror
equation (if we remember the equation!). It is however, a good
idea to imagine the situation graphically, before attempting the
formal solution. The image on a screen can only be formed in
front of the mirror. The image must be real. It is suggested in
the problem that the object is also real. We can draw a ray
diagram (backwards) to illustrate the situation.
I
F
O
d=5m
s'
2
s
Since we know the location of the screen, we should start our
analysis by drawing the image. The only ray we can track at
this time is the chief ray. We know that the object must be
somewhere along the object's chief ray. We have to look for a
point on the object's chief ray, which is five times closer to the
principle axis than the image is. From the similarity of the
triangles formed by the principle axis, the chief rays, the image
and the object, we can immediately figure out that the object
must be five times closer to the mirror than the image.
After we located the position of the object, we can track
both parallel rays, and graphically locate the focal point.
Let's now find the formal solution. In order to have a real
image enlarged five times, the magnification of the mirror in this
situation must be -5. We can either recall the relationship
between the magnification of a mirror (difficult to remember) or
consider the two similar triangles in the ray diagram formed by
the principal axis, the chief rays, the object and the image.
s'
s+d
(1)
m=− =−
s
s
b) This equation contains the only unknown requested in this
part of the problem. Solving for the position of the object we
find:
d
5m
s=−
=−
= 1. 25 m
m +1
( − 5) + 1
a) To describe the mirror we can use any characteristic of the
mirror. Let's for example find the focal length of the mirror. As
taken directly from the mirror equation
(2)
1 1 1
+ =
s s' f
the focal length of the mirror can be expressed as below
ss'
s(s + d )
1.25m ⋅ 6.25m
f=
= 1.04m
=
=
s + s' s + (s + d ) 1.25m + 6.25m
We can also obtain this result from geometrical considerations.
From the similarity of the triangles formed by the principle axis,
the object focal ray, the object and the mirror
1
s − f −h
(2’)
=
=
f
h' − m
Therefore
−5 ⋅ 125
ms
. m
f=
=
= 104
. m
−
1
−
5
−
1
m
(
) (
)
Positive focal length indicates a concave mirror.
(The curvature radius of the mirror is:
R = 2f = 2. 08 m )
4
Problem 36.21
A cubical block of ice 50 cm on a side is placed on a level floor over a speck of
dust. Find the location of the image of the speck if the index of refraction of ice
is 1.309.
β
n2
B
A
γ
n1
t’
dβ
t
I
α
dα
O
This is a case where an image is not really formed. The
refracted rays and their extensions do not meet at a single point.
We can see the speck of dust only because we observe the light
in a very small (differential) cone determined by the size of our
pupils. Only a small (differential) fragment (AB) of the surface
forms the observed image. (You have probably noticed in
nature, that in such an arrangement, the depth at which the
image is observed depends on the observation angle β.)
Although the text suggests solving the problem for a direct
from above observation, the solution presented here considers a
general case. In order to find the depth at which the image I is
observed one has to relate the dimensions of the (differential)
triangle ABI with the dimensions of the (differential) triangle
ABO. Both triangles have one angle which is very small
(differential size). From the sine theorem
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AB
OA
=
sin dα sin (180° − γ )
Geometrical considerations enable one to express the distance
OA in terms of the object’s depth t
OA =
t
cos α
and angle γ with the angle of incidence
γ = (90° + α + dα )
Expanding the trigonometric in Maclaurin series, one finds the
first significant terms to be
sin dα ≈ sin 0° + cos 0° ⋅ dα = dα
sin (180° − γ ) = sin (90° − (α + dα )) = cos(α + dα ) ≈ cos α
Solving the equations for AB
AB =
t
cos α
2
⋅ dα
Identical analysis of triangle ABI relates dimension AB with the
depth t’ of the image
AB =
t'
cos 2 β
⋅ dβ
Combining the last two equations eliminates AB and relates the
depths
cos 2 β dα
t' = t ⋅
⋅
2
cos α dβ
Using Snell’s law one can eliminate the dependence of the
image depth on one of the angles. Differentiation of Snell’s law
n1 sin α = n 2 sin β ,
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relates the differentials of the angles
n1 cos α ⋅ dα = n 2 cos β ⋅ dβ
and a direct transformation the angles themselves
1
⎛ ⎛ n ⎞2
⎞2
cos α = 1 − sin 2 α = ⎜1 − ⎜⎜ 2 ⎟⎟ sin 2 β ⎟
⎟
⎜ ⎝ n1 ⎠
⎝
⎠
Therefore
cos3 β n 2
cos3 β
n2
⋅
=
⋅
⋅
t' = t ⋅
t
3
2
cos3 α n1
⎛ ⎛n ⎞
⎞ 2 n1
⎜1 − ⎜ 2 ⎟ sin 2 β ⎟
⎜ ⎜⎝ n1 ⎟⎠
⎟
⎝
⎠
Note that the apparent depth t’ of the object has the highest
value when looked at directly from above (β=0°). In this case
t ' = 50cm ⋅
cos3 0°
⎛ ⎛ 1 ⎞
⎞
⎜1 − ⎜
sin 2 0° ⎟
⎟
⎜ ⎝ 1.309 ⎠
⎟
⎝
⎠
2
3
2
⋅
1
50cm
=
≈ 38.2cm
1.309 1.309
As the observation angle β increases, the image appears at a
shallower and shallower depth.
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Problem 36.33
The nickel’s image in Figure P36.31 has twice the diameter of the nickel and is
2.84 cm from the lens. Determine the focal length of the lens.
I
Though a magnifying glass one
looks at a virtual image. A
O
negative (opposite to the
distance from the lens) number
F
must therefore describe the
F
position
of
the
image.
Considering the chief ray, one
s
should recognize that in order
to have the magnification of m,
s’=-2.84 cm
the object must be placed
exactly at the midpoint between the lens and the image.
Therefore the object is at position
i
o
s=
− s'
m
The lens equation gives
⎛1 1 ⎞
f =⎜ + ⎟
⎝ s s' ⎠
−1
⎛ m 1⎞
= ⎜− + ⎟
⎝ s' s' ⎠
−1
⎛1 − m ⎞
=⎜
⎟
⎝ s' ⎠
−1
=
s'
− 2.84cm
=
= 2.84cm
1− m
1− 2
Note that an accurate ray diagram indicates that the object focal
point is at the location of the image. The position of the object
focal point at the same location of the image is coincidental and
results from this particular magnification.
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Problem 36.74
Two converging lenses with focal lengths of 10 cm and 20 cm are positioned
50 cm apart, as shown in Figure P36.74. The final image is to be located between
the lenses, at the position indicated. (a) How far to the left of the first lens should
the object be? (b) What is the overall magnification? (c) Is the final image upright
or inverted?
final
image
object
h
F1i
F2o
first
image
H
h'
p
q
s
l = 31 cm
s'
d = 50 cm
A ray diagram helps us to imagine the situation. Since the final
image (formed by the second lens) is not on the same side as the image
rays (refracted be the second lens) the lens forms a virtual image.
Tracing the chief ray and the image parallel ray (and the image focal
ray) allows one to determine the position of the object for the second
lens (image formed by the first lens). The found object (first image) is on
the same side as the object rays therefore the object is real. Tracing the
image chief, focal (and parallel) rays of the first lens allows one to
identify the initial object.
The algebraic solution can be found from the lens equation
applied to both lenses. Consistent with the figure
1)
1 1 1
+ =
p q f1
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2)
1 1 1
+ =
s s' f 2
Additionally
3)
q+s= d
4)
−s'+ l = d
We obtained four equations with four unknown. The rest is
algebra. We have to solve the above set for p. From the information
given (eq. 4) we can determine the position of the final image (with
respect to the second lens)
s' = −( d − l ) = −(50cm − 31cm ) = −19cm
We can use this information in equation 2 and find the position (with
respect to the second lens) of the first image
⎛ 1 1⎞
s=⎜ − ⎟
⎝ f 2 s'⎠
−1
1
1 ⎞
= ⎛⎜
−
⎟
⎝ 20cm -19cm⎠
−1
≈ 9.74cm
The position of the first image with respect to the first lens is therefore
(from eq. 3)
q = d − s ≈ 50cm − 9.74 ≈ 40.26cm
Now from equation 1
⎛ 1 1⎞
p=⎜ − ⎟
⎝ f2 q ⎠
−1
1 ⎞
⎛ 1
=⎜
−
⎟
⎝ 20cm 40.26cm⎠
−1
≈ 133
. cm
(It makes sense to perform calculation step-by-step. We can check each
step for possible discrepancies with the ray diagram.)
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b) Considering similar triangles, formed by the optical axis, the chief
rays, the objects and the images, one finds the following two
proportionalities
− h ' − s'
=
−H
s
and
−H q
=
h
p
From its definition, the magnification of the system is
M=
h'
− h ' − H s' q − 19.0cm 40.3cm
=−
⋅
= ⋅ =
⋅
≈ − 5. 9
h
−H h
s p 9.74cm 13.3cm
c) Negative magnification, as well as the ray diagram, indicates
that the final image is inverted.
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