18. Geometical Optics
351
L8.34 A parabolic mirror would produce a parallel
beam of light if the bulb is placed at the focus. This would be the best choice. A concave
spherical mirror would be the next best choice.
It would produce a parallel beam for paraxial
rays, but off-axis rays would not necessarily be
parallel to the paraxial rays as they emerge from
the mirror.
18.35 No. A convex mirror always forms a virtual
image and a concave mirror forms a virtual image if XQ < /, as is the case here.
18.36 Under water, the focal point of your eye increases since nwater is closer to niens.
18.40 A virtual image (— sign) is located 11 cm behind the window.
1 1
2
H
= -)— for a convex mirror, / = | is
negative. Here r =^"0.22 m and x^— 17 m, so
A
2
1 x -1
.22 m
17 m
-0.11 m = -ll cm
=
The image is virtual and 11 cm behind the mirror. The car is sufficiently distant as to be "at
oc."
r2
Thus your eye can no longer focus an image on
the retina. Putting on a mask returns nmea to
1.00.
18.37 They provide a larger field of view.
ray from distant object
'at large angle
18.42 — + — =
— for a convex mirror, since r is
r
negative, so
r|
is reflected toward eye
ZQ
For a real object, XQ is always positive, so that
Xi < 0, meaning that the image is virtual. To
get a real image XQ must be negative, i.e, the
the object must be virtual and
- r
18.38 The front side of the bubble acts like a convex
mirror and forms a virtual image. The back side
of the bubble acts like a concave mirror and
forms an inverted, real image. Rays of light
entering the camera after reflection from the
two bubble surfaces are focused by the camera
lens as if they had come from these two images.
Thus both images are recorded on the. film.
18.39
1
>
0>z0
0
> -^=-|/|
The object is closer than the focal point.
\
18.43jFor a spherical refracting surface use Equation
18.6
Here HI = 1.5,712 = 1.0, XQ = 6.0 cm and r —
—8.5 cm. So the wick appears to be at
— =- - -
! "I
Tl2 — Til
Here z0 = 1.0 m, and r — +0.50 m
-i
1
= 0.33 m
0.50 m 1.0 m
A real, inverted image, reduced in size, is formed
0.33 m from the mirror.
rs==""""7—-j
light bulb
0.50 m—*-
r
=
=
-0.5
-8.5 cm
-5.2 cm
1.0
air
1.0
1.5
6.0 cm
oil n — 1.5
• wick
,\ 6 cm
-i
Then ri = +12.5 cm, r2 = oo, and
1
(B
1 \
2.5 cm J
-!)!- 1, 0.57
12.5
o
"XT'
18.5(^Equation 18.4 gives the relation between the image and object coordinates:
_L
-L
I
X0
Xi
f
With a real object and real image, both coordinates are positive. Since the lens is convex
(and presumably used in air) then / is positive too. ^fe want to minimize the distance
3 = x0 + Xi. The quantity we can vary is the object distance, xa, so let's write 3 in terms
of x0 :
So for a real image, we must have xg > f. Then
._,.+„ _* + -fe__,,(szIi£)
Now we find the minimum:
da
2xg
dx0
x0- f
2-
Thus ds/dx0 = 0 when
or
Then the minimum separation is:
(2/)2
3mm =
We could check that this is a minimum by taking the second derivative. But since an infinite
separation is possible, that is clearly the maximum, so 4/ must be the minimum.
18.52 With a real image and a real object, both x0 and x< are positive, so d = x0 + X*.
8
18. Geometrical Optics
353
Using Equation 18.6 again, we have
1.50
1.00
+
0.62 m
1.00
=
(1.00 -1.50)
-0.50 m
+1.0
1.50
~m
0.62 m
-0.70 m
The image is virtual and is 0.70 m from the
outside surface of the window. When you
look through this window at the light bulb,
it appears to be further away than it actually
is. FVom the thin-lens approximation (Example
18.5) the image is 0.68 m from the center of the
lens, 3% smaller.
18.54 Reflection: Xi = —4.74 cm (virtual, inside the
ball's near surface.)
Refraction: 7.65 cm outside the ball's far
surface. The focal points are 15.0 cm from the
ball's center.
=
1.12 cm+2.00 cm = 3.12 cm
At the second surface,
Tii = 1.00,n2 = 1.42,7-2 = -1.00 cm
712 —
1.00
3.12 cm
=
Zij2
1.42
Xi2
0.42
-1.00 cm
1.00
-0.42
-1.42
1.00 cm
= -1.92 cm
i-l
3.12 cmj
The image is virtual and is located inside the
bubble. 0.92 cm from the center, on the same
side as the crystal. The bubble behaves like a
diverging lens.
18.56 5.4 cm below the air/oil surface.
18.57 The thickness of the lens is related to its diameter d and the radius of curvature of each
surface, r.
crystal
D
A
7x2
_ n-2 — n\
V
with Ti — +1.00 cm
3.00 cm
*i,i
=
=
+
d
-0.42
1.00 cm
-1.12 cm
(1-00)
2
DC = AC =
xiti '' +1.00 cm
1.42
3.00 cm
-i
Thus
= DC2
BC =
The first image is virtual. This images forms a
real object for the second surface.
And
t
=
= 2r
= 2(AC-BC)
Lea & Burke
358
Physics: The Nature of Things
18.66 / = -20 cm (2 sig. figs.). The focal lengti
is | /| = 20 cm. An erect virtual image of size
X. 22.0 cm is formed 12.0 cm from the lens.
±TJ!H
/^
V
-*~~-^v
IS.S'D For a lens in air, use Equation 18.10:
—^
! - f(n _ l)n f i
/~
(~i
incoming
light
18.63 For a convex mirror, R < 0, so
1
XQ
+
1 _ 2 _
2
Xi ~ R ~
\R\
1
~2
\
/
x0
Solving for Xi~.
Here n = 1.50 and r\ — —8.0 cm, r% = +16 cm
(see figure). So
\R\
XQ
-2*0-|*l
f
=
0.50 -
3.0 -
^ =
2
32
~~o~
Then the magnification is:
x0
1
\R\
|#| + 2xc
1
L8.64 Image is virtual, erect, and 5 the size of the
object.
18.65 Horizontal scale: 1 cm = 2 cm
Vertical scale: 1 cm = 0.5 cm.
Measured position of image 3.3 cm —> 6.6 cm
Measured size of image 0.5 cm —> 0.25 cm.
See Appendix 1 for figure.
=
•=
m
'1
I -r
16
cm
——H
CI
&
i y =f
/>
^_
1 M
=
3.0 _i_
* XQ)
\ 32 cm 16
32 cm
= —6.4 cm
^^ 5
XQ ~A 16 cm
' 2
-f- = >f 0.40
5
This is a virtual image, since Xi < 0.
T—yp
|-
J^bSL
1
+ 4.5 cm
V
l 16 cm/
For an object 16 cm from lens, XQ = 16 cm
which is always less then 1 for real objects
(x
\^O >
-^ 01
*-*/ "
Xi
8.0 cm
14 cmy
=
—6.63 cm = —n
14 cm
Size = 0.24 cm. This agrees with the figure—
the image is inverted and about ^ the size of
object.
Horizontal scale: 1 square = 4 cm
Vertical scale: not needed.
Image location Xi = —1.6 squares = —6.4 cm
Image is virtual and 0.4 the size of the object.
Lea & Burke
358
Thus the object must be virtual. Since the object viewed by the eyepiece is the image formed
by the objective lens, a real image is formed by
the eyepiece if the objective's image lies beyona
the eyepiece and inside /£.
So
li = 12.65 cm
Thus
12.65 cm
= 0.63
20 cm
This is consistent with the figure, given the accuracy with which we can measure.
c) For an arrow on axis, the image length would
be
— £ii - ar«2 = 12 cm
•
and the magnification would be
(slightly less).
objective image
in this region ,.
eyepiece
objective
= 0.60
18.72 Scale: 1 square = 5 cm
measured length of object = 1.5 cm in figure
measured length of image = 2.4 cm in figure
magnification = 0.48.
Physics; The Nature of Things
18.76 a) magnification preferred, orientation not important
b) image should be magnified and erect
c) image must be real, reduced hi size
d) magnification is important
e) image must be real and magnified
f) image should be virtual
g) image should be at oo. Other considerations
are irrelevant.
h) image should be erect, magnified, virtual
-—•^ i) image should be erect, virtual
^~T\
{ 18.77J1) Converging lens, first:
%>
*"———
Since object is at focal point of converging
lens;
-i
oo
18.73 Focal length has to be the same; / is not affected by flipping lens over. Each lens is converging, and so is the combination. The combination's focal length is half that of either lens
alone.
Diverging lens then "sees" object at x0 = oo.
So
/ I
1\
Xi = —
= —5.0 cm
V s
virtual image at position of 1st lens
2) Diverging lens first:
1
18.74 The eyeglasses move the image away from the
cornea.
18.75 The eyepiece is a diverging lens, so the image
distance is
_L .I
/£
-1
-1
= — — cm
2
This is object viewed by converging lens so object distance is
x0
5
15
= 5 cm+— cm = — cm
X0
with IE < 0 for a diverging lens. For this lens
to form a real image means that x^ > 0. This
means that — -^- is positive and greater than
j^r. Thus the object distance, x0, must be
f
oo;
<0
i
—
—
I
-—I
=
-i
cm
+15 cm
.'. real image 15 cm beyond second lens.