CHAPTER 25: Optical Instruments
Answers to Questions
1.
Stopping down a lens to a larger f-number means that the lens opening is smaller and only light rays
coming through the central part of the lens are accepted. These rays form smaller circles of confusion,
which means a greater range of object distances will be more sharply focused.
2.
If a lens is stopped down too much (too big of an f-number), then diffraction will occur as light passes
through the extremely small lens opening (the light will begin to “bend around” the edges of the stop).
This diffraction will blur the image, especially around the edges of the film, and will lead to an image
that is less sharp.
3.
The lens equation ⎜
⎛1 1 1⎞
= + ⎟ says that as an object gets closer to the lens ( d o decreases) and the
f
d o di ⎠
⎝
focal length of the lens remains constant, the image distance di must get larger to create a focused
image. Since we cannot move the film farther away from the lens, we need to move the lens farther
away from the film. This is unlike the human eye, where the focal length of the lens is changed as the
object distance changes.
4.
As people get older, their eyes can no longer accommodate as well. It becomes harder for the muscles to
change the shape of the lens, since the lens becomes less flexible with age. As people get older, their
near point increases and becomes greater than the ideal value of 25 cm. They may still need the
“regular” upper portion of their lenses so they can have a far point at infinity (most people need these at
much younger ages), but now an older person will need a bifocal in the lower portion of their lenses to
create a near point back at 25 cm for seeing close objects. Thus, as people get older, their far point is too
small and their near point is too large. Bifocals can correct both of these problems.
5.
No, a nearsighted person will not be able to
in air
see clearly if they wear their corrective
lenses underwater. A nearsighted person
has a far point that is closer than infinity
and they wear corrective lenses to bring the
image of a far away object to their far point
so they can see it clearly. See the first pair
of diagrams. The object is at infinity. In
in water
air, the image is at the far point. If the
person’s eyes and glasses are underwater,
and since the index of refraction of glass is
closer to that of water than to that of air,
the glasses will not bend the light as much
as they did in the air. Therefore, the image
of the faraway object will now be at a position that is beyond the person’s far point.
The image will now be out of focus.
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
230
Physics: Principles with Applications, 6th Edition
Giancoli
6.
If a person’s face appears narrower through the
person’s glasses, the lenses are diverging. If a
person is wearing diverging lenses, then they are
nearsighted. Since the rays from a person’s ears
will be bent away from the optical axis when they
go through a diverging corrective lens, when that
ray gets to our eyes, our brain follows that ray
back through the lens without bending it (our
brain doesn’t know that light can bend) and
places the ears closer to where the person’s eyes
actually are. Thus, diverging corrective lenses
(for nearsighted people) make a person’s head
appear narrower when viewed through their glasses by an outside observer. In the diagram, a person
viewing from the left would see an image of the ears about where the eyes should be.
7.
To see far objects clearly, you want your eye muscles relaxed, which makes your lens relatively flat
(large focal length). When you f-stop your eye down by closing your eyelids partially (squinting), you
are only using the middle of your lens, where it is the most flat. This creates a smaller circle of
confusion for the lens, which helps you see distant objects more clearly.
8.
The images formed on our retinas of objects we look at are inverted and real. The implication of this is
that our brains must flip this inverted image for us so that we can see things upright.
9.
You can leave your eyes open while you are moving your head and still see clearly because your brain
“refreshes” the image from the retina about 30 times a second. In other words, your brain and your
retina work together in a manner similar to how a motion picture camera and film work together, which
is not at all like how a still camera and film work together.
10. Yes, reading glasses are magnifiers. Similarity: they are both converging lenses. Difference: glasses
try to put the image of the object at a person’s near point so it is clear to see, while a magnifying lens
tries to put the image of the object at infinity so it is incredibly large and it subtends as large of an angle
as possible on our retina. Thus, the main difference is that the glasses need to have the object closer to
the lens than the focal point, while the magnifying lens needs to have the object directly at the focal
point.
11. A poor quality, inexpensive lens will not be correctly shaped to fix chromatic aberrations. Thus, the
colors you see around the edges of these lenses are from all of the different colors of light getting
focused at different points, instead of all being focused at the same point.
12. To minimize spherical aberrations of a planoconvex lens used to
form a real image of an object at infinity, the convex side should
face the object. In this situation, both surfaces bend the
incoming parallel light rays a small amount each and nearly an
equal amount each, which reduces the spherical aberrations. If
the planar surface faced the object, then all of the bending of the
light rays would take place only on the convex side where the
rays emerge from the lens. Thus, with no bending on one face
and all of the bending on the other face, the amount of bend is
very different on the two faces and spherical aberrations will
cause severe problems.
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
231
Chapter 25
Optical Instruments
13. Spherical aberrations can be present in a simple lens. To correct this in a simple lens, usually many
lenses are used together in combination to minimize the bending at each of the surfaces. Your eye
minimizes spherical aberrations by also bending the light at many different interfaces as it makes its way
through the different parts of the eye (cornea, aqueous humor, lens, vitreous humor, etc.), each with their
own n. Also, the cornea is less curved at the edges than it is at the center and the lens is less dense at the
edges than at the center. Both of these reduce spherical aberrations in the eye since they cause the rays at
the outer edges to be bent less strongly. Curvature of field occurs when the focal plane is not flat. Our
curved retina helps with this distortion, whereas a flat piece of film in a camera, for example, wouldn’t
be able to fix this. Distortion is a result of the variation of the magnification at different distances from
the optical axis. This is most common in wide-angle lenses, where it must be corrected for. This is
compensated in the human eye because it is a very small lens and our retina is curved. Chromatic
aberrations are mostly compensated for in the human eye because the lens absorbs shorter wavelengths
and the retina is not very sensitive to most blue and violet wavelengths where most chromatic aberrations
occur.
14. Chromatic aberrations in lenses occur due to dispersion, where different colors (or wavelengths) of light
are bent different amounts due to the fact that the index of refraction of most materials varies with
wavelength. So, when light goes through a lens (a curved transparent material), not all of the different
colors come out of the lens at the exact same angle, but they are all focused at different positions. A
mirror, in contrast, reflects light off of a smooth metallic surface. This surface reflects all different colors
of light at the exact same angle, thus there is no refraction, no dispersion and no chromatic aberrations.
Of course, in most mirrors, there is a piece of glass that covers the metallic reflector. Since the two faces
of this piece of glass are parallel to each other, even though refraction and dispersion take place inside
the piece of glass, when the light emerges from the parallel face, all of the different colors are once again
going in the same direction, and there is no dispersion and no chromatic aberration.
15. The Rayleigh criterion gives us the resolution limit for two objects a distance D apart when using light of
1.22λ
, which can be interpreted as the smaller the angle, the better the resolution.
D
1.22 ( 450 nm )
1.22 ( 700 nm )
and θ red =
, we have
Looking at the two wavelengths given, θ blue =
D
D
1.22 ( 450 nm )
θ blue
450
D
=
=
= 0.64 . This could be expressed as saying that the resolution with blue
θ red 1.22 ( 700 nm ) 700
D
1
light is
= 1.56 times the resolution with red light.
0.64
wavelength λ: θ =
16. A large reflecting mirror in a telescope has several advantages over a large refracting lens: One, the
construction and grinding of a huge lens is very difficult (a mirror only needs one side to be ground and
smoothed to perfection). Two, a huge lens is so heavy that it will sag under its own weight, especially
since it needs to be held only along the edges of the lens in the telescope (mirrors can be well-supported
along their entire back, non-reflecting, side). Three, a huge lens will have serious chromatic aberrations
(a mirror has no chromatic aberrations).
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
232
Physics: Principles with Applications, 6th Edition
Giancoli
17. To get the best resolution from a microscope, you should use blue/violet light. The Rayleigh criterion
1.22λ ⎞
⎛
⎜θ =
⎟ shows that the shorter wavelengths of light have higher resolution. (Of course, you would
D ⎠
⎝
need very special lenses, since most glass absorbs blue/violet light very strongly.)
18. No, visible light cannot be used to “see” individual atoms. It is not possible to resolve the detail of
objects smaller than the wavelength of the radiation being used to look at the object. The shortest
wavelength of visible light is approximately 400 nm = 4 × 10−5 m , which is too big to use to see atoms
that are ∼ 10−8 m in diameter. Looking at the Rayleigh criterion, it gives:
−5
1.22λ 1.22 ( 4 ×10 m )
=
= 4880 rad . This is an impossible value, since our eyes have a value of
D
10−8 m
about 5 ×10−4 rad and microscopes can only improve this by about 1000X.
θ=
Solutions to Problems
1. From the definition of the f-stop, we have
f
f -stop = ;
D
( 55mm ) , which gives D = 39 mm;
1.4 =
1.4
D1.4
22 =
( 55mm ) ,
D22
which gives D22 = 2.5mm.
Thus the range of diameters is 2.5mm ≤ D ≤ 39 mm.
2. We find the f-number from
(14cm )
f
f
f -stop = =
=
.
D ( 6.0cm ) 2.3
3. The exposure is proportional to the area and the time:
Exposure ∝ At ∝ D 2t ∝ t
.
2
( f -stop )
Because we want the exposure to be the same, we have
t1
t2
=
;
2
2
( f -stop1 ) ( f -stop 2 )
⎡⎛ 1 ⎞ ⎤
⎢⎜ 250 ⎟ s ⎥
t
⎛ 1 ⎞
⎠ ⎦
⎣⎝
= 2 2 , which gives t2 = ⎜ ⎟ s.
2
⎝ 60 ⎠
( 5.6 )
(11)
4. The exposure is proportional to the area and the time:
Exposure ∝ At ∝ D 2t ∝ t
.
2
( f -stop )
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
233
Chapter 25
Optical Instruments
Because we want the exposure to be the same, we have
t1
t2
=
;
2
2
( f -stop1 ) ( f -stop 2 )
⎡⎛ 1 ⎞ ⎤ ⎡⎛ 1 ⎞ ⎤
⎢⎜ 125 ⎟ s ⎥ ⎢⎜ 1000 ⎟ s ⎥
f
⎠ ⎦ ⎣⎝
⎠ ⎦
⎣⎝
=
, which gives f -stop 2 = .
2
2
4
(11)
( f -stop2 )
5. For an object at infinity, the image will be in the focal plane, so we have d1 = f = 135mm.
When the object is at 1.2 m, we locate the image from
⎛ 1 ⎞ ⎛1⎞ 1
⎜ ⎟ +⎜ ⎟= ;
⎝ d o ⎠ ⎝ di ⎠ f
⎛
⎞ ⎛1⎞
1
1
, which gives di = 152 mm.
⎜
⎟ +⎜ ⎟=
⎝ 1200 mm ⎠ ⎝ d i ⎠ 135cm
Thus the distance from the lens to the film must change by
d i − f = 152 mm − 135mm = 17 mm.
6. We find the object distances from
⎛ 1 ⎞ ⎛1⎞ 1
⎜ ⎟ +⎜ ⎟= ;
⎝ d o ⎠ ⎝ di ⎠ f
⎛ 1 ⎞ ⎛
⎞
1
1
, which gives d o1 = ∞;
⎜
⎟+⎜
⎟=
⎝ d o1 ⎠ ⎝ 200.0 mm ⎠ 200 mm
⎛ 1 ⎞ ⎛
⎞
1
1
, which gives d o1 = 6.87 × 103 mm = 6.87 m.
⎜
⎟+⎜
⎟=
d
200.6
mm
200
mm
⎠
⎝ o2 ⎠ ⎝
Thus the range of object distances is 6.87 m ≤ d o ≤ ∞.
7. The converging camera lens will form a real, inverted image. For the magnification, we have
h
d
m= i =− i ;
ho
do
( 24 × 10
−
−3
( 28m )
m)
=−
di
, or d i = 4.97 × 10−2 m.
( 58m )
We find the focal length of the lens from
⎛ 1 ⎞ ⎛1⎞ 1
⎜ ⎟ + ⎜ ⎟= ;
⎝ d o ⎠ ⎝ di ⎠ f
⎛ 1 ⎞ ⎛ 1
⎞ 1
× 10−2 m ⎟ = , which gives f = 5.0 × 10−2 m = 50 mm.
⎜ m⎟ + ⎜
50
4.97
⎝
⎠ ⎝
⎠ f
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
234
Physics: Principles with Applications, 6th Edition
Giancoli
8. From the similar triangles on the ray diagram, we see that
H h
= ;
L1 L2
( 2.0cm ) = h
,
(100cm ) ( 7.0cm )
which gives h = 0.014cm = 1.4 mm.
To find the radius of each spot of the image, we consider the
light going through a slit and find the distance from the
central bright spot to the first dark spot. For destructive
interference, the path-length difference to the first dark spot is
d sin θ = !λ ;
We find the locations on the screen from
y = L2 tan θ .
For small angles, we have
sin θ ≈ tan θ , which gives
h
H
L1
L2
−2
−9
⎛ !λ ⎞ ! L λ ( 7.0 × 10 m )( 550 × 10 m )
y = L2 ⎜ ⎟ = 2 =
= 1.9 × 10−5 m = 0.019 mm.
−3
d
d
1.0
×
10
m
⎝ ⎠
(
)
Thus the diameter of the image spot is about 0.04 mm, which is much smaller than the separation of the
spots, so they are easily resolvable.
9. We find the effective f-number for the pinhole:
f ( 70 mm )
f
f -stop 2 = =
= .
D (1.0 mm ) 70
The exposure is proportional to the area and the time:
Exposure ∝ At ∝ D 2t ∝ t
.
2
( f -stop )
Because we want the exposure to be the same, we have
t1
t2
=
;
2
2
( f -stop1 ) ( f -stop2 )
⎡⎛ 1 ⎞ ⎤
⎢⎜ 250 ⎟ s ⎥
t
⎛1⎞
⎠ ⎦
⎣⎝
= 2 2 , which gives t2 = ⎜ ⎟ s.
2
⎝6⎠
(11)
( 70 )
10. The length of the eyeball is the image distance for a far object, i.e., the focal length of the lens.
We find the f-number from
f ( 20 mm )
f
.
f -stop = =
= 2.5 or
2.5
D ( 8.0 mm )
11. With the contact lens, an object at infinity would have a virtual image at the far point of the eye.
We find the power of the lens from
⎛ 1 ⎞ ⎛1⎞ 1
⎜ ⎟ + ⎜ ⎟ = = P, when distances are in m;
⎝ d o ⎠ ⎝ di ⎠ f
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
235
Chapter 25
Optical Instruments
⎞
1
⎛1⎞ ⎛
⎟ = P = − 5.9 D.
⎜ ⎟+⎜
⎝ ∞ ⎠ ⎝ − 0.17 m ⎠
To find the new near point, we have
⎛ 1 ⎞ ⎛1⎞ 1
⎜ ⎟ + ⎜ ⎟ = = P;
⎝ d o ⎠ ⎝ di ⎠ f
⎛ 1 ⎞ ⎛
⎞
1
⎜ ⎟+⎜
⎟ = − 5.9 D, which gives d o = 0.41m.
⎝ d o ⎠ ⎝ − 0.12 m ⎠
Glasses would be better, because they give a near point of 32 cm from the eye.
12. With the lens, the screen placed 55 cm from the eye, or 53.2 cm from the lens, is to produce a virtual
image 115 cm from the eye, or 113.2 cm from the lens.
We find the power of the lens from
⎛ 1 ⎞ ⎛1⎞ 1
⎜ ⎟ + ⎜ ⎟ = = P, when distances are in m;
⎝ d o ⎠ ⎝ di ⎠ f
⎛
⎞ ⎛
⎞
1
1
⎜
⎟+⎜
⎟ = P = + 1.0 D.
⎝ 0.532 m ⎠ ⎝ −1.132 m ⎠
13. With the glasses, an object at infinity would have its image
14 cm from the eye or 14cm − 2cm = 12cm from the lens; di = −12cm.
⎞
1 ⎛ 1 ⎞ ⎛1⎞ ⎛1⎞ ⎛
1
=⎜ ⎟ + ⎜ ⎟=⎜ ⎟ + ⎜
⎟ = − 8.3D
f ⎝ d o ⎠ ⎝ d i ⎠ ⎝ ∞ ⎠ ⎝ − 0.12 m ⎠
With the contact lenses, d i = − 14 cm.
P=
P=
⎞
1 ⎛ 1 ⎞ ⎛1⎞ ⎛1⎞ ⎛
1
=⎜ ⎟ + ⎜ ⎟=⎜ ⎟ + ⎜
⎟ = − 7.1D
f ⎝ d o ⎠ ⎝ d i ⎠ ⎝ ∞ ⎠ ⎝ − 0.14 m ⎠
14. The actual near point of the person is 45 cm. With the lens, an object placed at the normal near point,
25 cm, or 23 cm from the lens, is to produce a virtual image 45 cm from the eye, or 43 cm from the lens.
We find the power of the lens from
⎛ 1 ⎞ ⎛1⎞ 1
⎜ ⎟ + ⎜ ⎟ = = P, when distances are in m;
⎝ d o ⎠ ⎝ di ⎠ f
⎛ 1 ⎞ ⎛
⎞
1
⎜
⎟+⎜
⎟ = P = + 2.0 D.
⎝ 0.23m ⎠ ⎝ − 0.43m ⎠
15. (a) Since the diopter is negative, the lens is diverging, so it produces images closer than the object; thus
the person is nearsighted.
(b) We find the far point by finding the image distance for an object at infinity:
⎛ 1 ⎞ ⎛1⎞ 1
⎜ ⎟ + ⎜ ⎟ = = P;
⎝ d o ⎠ ⎝ di ⎠ f
⎛1⎞ ⎛1⎞
⎜ ⎟ + ⎜ ⎟ = − 3.50 D, which gives di = − 0.286 m = − 28.6cm.
⎝ ∞ ⎠ ⎝ di ⎠
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
236
Physics: Principles with Applications, 6th Edition
Giancoli
Because this is the distance from the lens, the far point without glasses is
28.6cm + 2.0cm = 30.6cm.
16. (a) We find the power of the lens for an object at infinity:
⎛ 1 ⎞ ⎛1⎞ 1
⎜ ⎟ + ⎜ ⎟ = = P;
⎝ d o ⎠ ⎝ di ⎠ f
⎞
1
⎛1⎞ ⎛
⎟ = P = −1.3D.
⎜ ⎟+⎜
⎝ ∞ ⎠ ⎝ − 0.75m ⎠
(b) To find the near point with the lens, we have
⎛ 1 ⎞ ⎛1⎞ 1
⎜ ⎟ + ⎜ ⎟ = = P;
⎝ d o ⎠ ⎝ di ⎠ f
⎛ 1
⎜
⎝ do
⎞ ⎛
⎞
1
⎟+⎜
⎟ = −1.3D, which gives d o = 0.37 m = 37 cm.
⎠ ⎝ − 0.25m ⎠
17. The 2.0 cm of a normal eye is the image distance for an object at infinity; thus it is the focal length of the
lens of the eye. To find the length of the nearsighted eye, we find the image distance (distance from lens
to retina) for an object at the far point of the eye:
⎛ 1 ⎞ ⎛1⎞ 1
⎜ ⎟ +⎜ ⎟= ;
⎝ d o ⎠ ⎝ di ⎠ f
⎛ 1 ⎞ ⎛1⎞
1
, which gives di = 2.27 cm.
⎜
⎟+⎜ ⎟=
⎝ 17 cm ⎠ ⎝ d i ⎠ 2.0cm
Thus the difference is 2.27 cm − 2.0cm = 0.3cm.
18. We find the far point of the eye by finding the image distance with the lens for an object at infinity:
⎛ 1 ⎞ ⎛ 1 ⎞ 1
⎜
⎟ +⎜ ⎟= ;
⎝ d o1 ⎠ ⎝ di1 ⎠ f1
1
⎛1⎞ ⎛ 1 ⎞
, which gives di1 = − 22.0cm from the lens, or 23.8 cm from the eye.
⎜ ⎟ + ⎜ ⎟=
⎝ ∞ ⎠ ⎝ di1 ⎠ − 22.0cm
We find the focal length of the contact lens from
⎛ 1 ⎞ ⎛ 1 ⎞ 1
⎜
⎟+⎜
⎟= ;
⎝ d o2 ⎠ ⎝ di2 ⎠ f 2
⎞ 1
1
⎛1⎞ ⎛
⎟ = , which gives f 2 = − 23.8cm.
⎜ ⎟+⎜
⎝ ∞ ⎠ ⎝ − 23.8cm ⎠ f 2
19. (a) We find the focal length of the lens for an object at infinity and the image on the retina:
⎛ 1 ⎞ ⎛ 1 ⎞ 1
⎜
⎟ +⎜ ⎟= ;
⎝ d o1 ⎠ ⎝ di1 ⎠ f1
⎛1⎞ ⎛ 1 ⎞ 1
⎟ = , which gives f1 = 2.0cm.
⎜ ⎟+⎜
⎝ ∞ ⎠ ⎝ 2.0cm ⎠ f1
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
237
Chapter 25
Optical Instruments
(b) With the object 30 cm from the eye, we have
⎛ 1 ⎞ ⎛ 1 ⎞ 1
⎜
⎟ +⎜ ⎟= ;
⎝ d o2 ⎠ ⎝ di2 ⎠ f 2
⎛ 1 ⎞ ⎛ 1 ⎞ 1
⎜
⎟+⎜
⎟ = , which gives f 2 = 1.9cm.
⎝ 33cm ⎠ ⎝ 2.0cm ⎠ f 2
20. We find the object distance for an image at her eye’s near point:
⎛ 1 ⎞ ⎛1⎞ 1
⎜ ⎟ + ⎜ ⎟ = = P;
⎝ d o ⎠ ⎝ di ⎠ f
⎛ 1 ⎞ ⎛
⎞
1
−2
⎜ ⎟+⎜
⎟ = − 4.0 D, which gives d o = 1.7 × 10 m = 17 cm.
d
−
0.100
m
⎠
⎝ o⎠ ⎝
We find the object distance for an image at her eye’s far point:
⎛ 1 ⎞ ⎛1⎞ 1
⎜ ⎟ + ⎜ ⎟ = = P;
⎝ d o ⎠ ⎝ di ⎠ f
⎛ 1 ⎞ ⎛
⎞
1
⎜ ⎟+⎜
⎟ = − 4.0 D, which gives d o = 1.0 m = 100cm.
⎝ d o ⎠ ⎝ − 0.200 m ⎠
21. The magnification with the image at infinity is
N ( 25cm )
M= =
= 2.1 × .
f (12cm )
22. We find the focal length from
N
M= ;
f
3.5 =
( 25cm ) ,
f
which gives f = 7.1cm.
23. (a) We find the focal length with the image at the near point from
N
M =1+ ;
f1
2.5 =
1 + ( 25cm )
, which gives f1 = 17 cm.
f1
(b) If the eye is relaxed, the image is at infinity, so we have
N
M= ;
f2
2.5 =
( 25cm ) , which gives
f2
f 2 = 10cm.
24. Maximum magnification is obtained with the image at the near point. We find the object distance from
⎛ 1 ⎞ ⎛1⎞ 1
⎜ ⎟ +⎜ ⎟= ;
⎝ d o ⎠ ⎝ di ⎠ f
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
238
Physics: Principles with Applications, 6th Edition
Giancoli
⎛ 1 ⎞ ⎛
⎞
1
1
cm, which gives d o = 6.6 cm from the lens.
⎜ ⎟+⎜
⎟=
⎝ d o ⎠ ⎝ − 25.0cm ⎠ 9.00
The magnification is
( 25.0cm )
N
M =1 +
=1 +
= 3.78 × .
f
( 9.00cm )
25. (a) The angular magnification with the image at the near point is
( 25.0cm )
N
M =1 +
=1 +
= 3.63 × .
f
( 9.50cm )
hi
(b) Because the object without the lens and the image with the
lens are at the near point, the angular magnification is also
the ratio of widths:
h
M= i;
ho
hi
3.63 =
, which gives hi = 12.0 mm.
3.30
( mm )
θ′
ho
O
I
N
ho
θ
O
(c) We find the object distance from
⎛ 1 ⎞ ⎛1⎞ 1
⎜ ⎟ +⎜ ⎟= ;
⎝ d o ⎠ ⎝ di ⎠ f
⎛ 1
⎜
⎝ do
which gives
⎞ ⎛
⎞
1
1
cm,
⎟+⎜
⎟=
⎠ ⎝ − 25.0cm ⎠ 9.50
d o = 6.88 cm from the lens.
26. (a) We find the image distance from
⎛ 1 ⎞ ⎛1⎞ 1
⎜ ⎟ +⎜ ⎟= ;
⎝ d o ⎠ ⎝ di ⎠ f
⎛ 1 ⎞ ⎛1
⎜
⎟+⎜
⎝ 5.55cm ⎠ ⎝ d i
hi
⎞
1
cm,
⎟=
⎠ 6.00
θ′
ho
O
I
which gives di = − 74cm.
N
(b) From the diagram we see that the angular
magnification is
⎛ ho ⎞
⎜ ⎟
d
θ′
N
M = =⎝ o⎠=
h
θ ⎛ o ⎞ do
⎜N⎟
⎝ ⎠
25cm
(
)
=
= 4.50 × .
( 5.55cm )
ho
θ
O
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
239
Chapter 25
Optical Instruments
27. (a) We find the image distance from
⎛ 1 ⎞ ⎛1⎞ 1
⎜ ⎟ +⎜ ⎟= ;
⎝ d o ⎠ ⎝ di ⎠ f
⎛ 1 ⎞ ⎛1⎞ 1
cm, which gives di = − 64cm.
⎜
⎟ +⎜ ⎟=
⎝ 7.5cm ⎠ ⎝ di ⎠ 8.5
(b) The angular magnification is
⎛ ho ⎞
⎜ ⎟
θ ′ ⎝ do ⎠ N
M= =
=
θ ⎛ ho ⎞ d o
⎜N⎟
⎝ ⎠
=
( 25cm )
=
( 7.5cm )
3.3 × .
28. We find the focal length of the lens from
N
M= ;
f
3.0 =
( 25cm ) ,
f
which gives f = 8.3cm.
(a) The magnification with the image at infinity is
( 55cm )
N
M1 = 1 =
= 6.6 × .
f
( 8.3cm )
(b) The magnification with the image at infinity is
(16cm )
N
= 1.9 × .
M2 = 2 =
f
(8.3cm )
Without the lens, the closest an object can be placed is the near point. A farther near point means a
smaller angle subtended by the object without the lens, and thus greater magnification.
29. The magnification of the telescope is given by
( 76cm )
f
M =− o =−
= − 27 × .
fe
( 2.8cm )
For both object and image far away, the separation of the lenses is
L = f o + f e = 76cm + 2.8cm = 79cm.
30. We find the focal length of the eyepiece from the magnification:
f
M =− o;
fe
− 25 = −
( 78cm ) ,
which gives f e = 3.1cm.
fe
For both object and image far away, the separation of the lenses is
L = f o + f e = 78cm + 3.1cm = 81cm.
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
240
Physics: Principles with Applications, 6th Edition
Giancoli
31. We find the focal length of the objective from the magnification:
f
M = o;
fe
fo
8.0 =
, which gives f o = 22cm.
( 2.8cm )
32. We find the focal length of the eyepiece from the power:
1 1
fe = =
D = 0.029 m = 2.9cm.
P 35
The magnification of the telescope is given by
(85cm )
f
M =− o =−
= − 29 × .
fe
( 2.9cm )
33. For both object and image far away, we find the focal length of the eyepiece from the separation of the
lenses:
L = fo + fe ;
75.2cm = 74.5cm + f e , which gives f e = 0.7 cm.
The magnification of the telescope is given by
( 75.2cm )
f
M =− o =−
= −110 × .
fe
( 0.7 cm )
34. For both object and image far away, we find the (negative) focal length of the eyepiece from the
separation of the lenses:
L = fo + fe ;
32.8cm = 36.0cm + f e , which gives f e = − 3.2cm.
The magnification of the telescope is given by
( 36.0cm )
f
M =− o =−
= 11 × .
fe
( − 3.2cm )
35. The reflecting mirror acts as the objective, with a focal length
r ( 6.0 m )
fo = =
= 3.0 m.
2
2
The magnification of the telescope is given by
( 300cm )
f
M =− o =−
= − 94 × .
fe
( 3.2cm )
36. We find the focal length of the mirror from
f
M =− o;
fe
fo
−120 = −
, which gives f o = 3.8 × 102 cm = 3.8m.
( 3.2cm )
The radius is
r = 2 f o = 2 ( 3.8m ) = 7.6 m.
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
241
Chapter 25
Optical Instruments
37. For the magnification we have
f
M = − o = −170, or f o = 170 f e .
fe
For both object and image far away, we have
L = fo + fe ;
1.25m = 170 f e + f e , which gives f e = 7.31 × 10−3 m = 7.31mm.
The focal length of the objective is
f o = 170 f e = 170 ( 7.31 × 10−3 m ) = 1.24 m.
38. First find the image distance of the star from the objective mirror. The star is off at
infinity, so this will be at the focal point to the left of the objective mirror.
r 3.0 m
d1i = f1 = 1 =
= 1.5m
2
2
Then the object distance of the image of the star from the eyepiece mirror is
0.9 m − 1.5m = − 0.6 m
The negative sign indicates the direction to the left of the eyepiece mirror.
Next find the image distance of the image of the star from the eyepiece mirror.
1
1
1
+
=
d 2i d 2o f 2
1
1
1
1
1
1
1
= −
=
−
=
−
d 2i f 2 d 2o ⎛ r2 ⎞ d 2o ( − 0.75m ) ( − 0.6 m )
⎜ ⎟
⎝2⎠
Solve for d 2i to get
d 2i = 3m.
39. The magnification of the microscope is
( 25cm )(17.5cm )
N
M= l =
= 480 × .
f o f e ( 0.65cm )(1.40cm )
40. We find the focal length of the eyepiece from the magnification of the microscope:
N
M= l ;
fo fe
620 =
( 25cm )(17.5cm )
,
( 0.40cm ) fe
which gives f e = 1.8cm.
41. The total magnification is
( 25cm )(17 cm )
N
M= l =
= 610 × .
f e f o ( 2.5cm )( 0.28cm )
42. (a) The total magnification is
M = M o M e = ( 59.0 )(12.0 ) = 708 × .
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
242
Physics: Principles with Applications, 6th Edition
Giancoli
(b) With the final image at infinity, we find the focal length of the eyepiece from
N
Me = ;
fe
12.0 =
( 25.0cm ) ,
59.0 =
(17.9cm ) ,
which gives f e = 2.08cm.
fe
Because the image from the objective is at the focal point of the eyepiece, the image distance for
the objective is
di = l − f e = 20.0cm − 2.08cm = 17.9cm.
We find the object distance from the magnification:
d
Mo = i ;
do
which gives d o = 0.303cm.
do
We find the focal length of the objective from the lens equation:
⎛ 1 ⎞ ⎛1⎞ 1
⎜ ⎟+⎜ ⎟= ;
⎝ d o ⎠ ⎝ di ⎠ f o
⎛
⎞ ⎛ 1 ⎞ 1
1
⎜
⎟+⎜
⎟ = , which gives f o = 0.298cm.
⎝ 0.289cm ⎠ ⎝ 17.9cm ⎠ f o
(c) We found the object distance: d o = 0.303cm.
43. (a) Because the image from the objective is at the focal point of the eyepiece, the image distance for
the objective is
dio = l − f e = 16.0cm − 1.8cm = 14.2cm.
We find the object distance from the lens equation for the objective:
⎛ 1 ⎞ ⎛ 1 ⎞ 1
⎜
⎟+⎜
⎟= ;
⎝ d oo ⎠ ⎝ dio ⎠ f o
⎛ 1 ⎞ ⎛ 1 ⎞
1
, which gives d oo = 0.85cm.
⎜
⎟+⎜
⎟=
⎝ d oo ⎠ ⎝ 14.2 cm ⎠ 0.80cm
(b) With the final image at infinity, the magnification of the eyepiece is
N ( 25.0cm )
Me = =
= 13.9 × .
f
(1.8cm )
The magnification of the objective is
(14.2cm )
d
M o = io =
= 16.7 × .
d oo ( 0.85cm )
The total magnification is
M = M o M e = (16.7 )(13.9 ) = 230 × .
44. (a) We find the object distance from the lens equation for the eyepiece:
⎛ 1 ⎞ ⎛ 1 ⎞ 1
⎜
⎟ +⎜ ⎟= ;
⎝ d oe ⎠ ⎝ die ⎠ f e
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
243
Chapter 25
Optical Instruments
⎛ 1 ⎞ ⎛ 1 ⎞
1
, which gives d oe = 1.7 cm.
⎜
⎟+⎜
⎟=
⎝ d oe ⎠ ⎝ − 25 cm ⎠ 1.8cm
The image distance for the objective is
dio = l − d oe = 16.0cm − 1.7 cm = 14.3cm.
We find the object distance from the lens equation for the objective:
⎛ 1 ⎞ ⎛ 1 ⎞ 1
⎜
⎟+⎜
⎟= ;
⎝ d oo ⎠ ⎝ d io ⎠ f o
⎛ 1 ⎞ ⎛ 1 ⎞
1
, which gives d oo = 0.85cm.
⎜
⎟+⎜
⎟=
⎝ d oo ⎠ ⎝ 14.3cm ⎠ 0.80cm
(b) With the final image at the near point, the magnification of the eyepiece is
( 25.0 cm )
N
Me =1 +
=1 +
= 14.9 × .
f
(1.8 cm )
The magnification of the objective is
(14.3cm )
d
M o = io =
= 16.9 × .
doo ( 0.85cm )
The total magnification is
M = M o M e = (16.9 )(14.9 ) = 250 × .
45. (a) We find the image distance from the lens equation for the objective:
⎛ 1 ⎞ ⎛ 1 ⎞ 1
⎜
⎟+⎜
⎟= ;
⎝ d oo ⎠ ⎝ d io ⎠ f o
⎛
⎞ ⎛ 1 ⎞
1
1
cm, which gives dio = 11.7 cm.
⎟=
⎜
⎟+⎜
⎝ 0.790cm ⎠ ⎝ d io ⎠ 0.740
For the relaxed eye, the image from the objective is at the focal point of the eyepiece: d oe = 2.70cm.
The distance between lenses is
l = dio + d oe = 11.7 cm − 2.70cm = 14.4cm.
(b) With the final image at infinity, the magnification of the eyepiece is
N ( 25.0cm )
Me = =
= 9.26 × .
f ( 2.70cm )
The magnification of the objective is
(11.7 cm )
d
M o = io =
= 14.8 × .
d oo ( 0.790cm )
The total magnification is
M = M o M e = (14.8 )( 9.26 ) = 137 × .
46. (a) When lenses are in contact, the powers add:
⎛
⎞ ⎛ 1 ⎞
1
P = P1 + P2 = ⎜
⎟+⎜
⎟ = + 0.776 D.
⎝ − 0.28m ⎠ ⎝ 0.23m ⎠
It is a positive lens, and thus converging.
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
244
Physics: Principles with Applications, 6th Edition
Giancoli
(b) The focal length is
1
1
f = =
= 1.3m.
P 0.776 D
47. (a) We find the incident angle from
(1.0cm )
h
= 0.0833, so θ1 = 4.78°.
sin θ1 = 1 =
R (12.0cm )
For the refraction at the curved surface, we have
sin θ1 = n sin θ 2 ;
sin 4.78° = (1.50 ) sin θ 2 ,
θ1 – θ2 = θ3
θ1
which gives sin θ 2 = 0.0556, so θ 2 = 3.18°.
We see from the diagram that
θ 3 = θ1 − θ 2 = 4.78° − 3.18° = 1.60°.
For the refraction at the flat face, we have
n sin θ 3 = sin θ 4 ;
θ2
h
θ3
θ1
θ4
θ4
d
n
(1.50 ) sin1.60° = sin θ 4 ,
h′
R
which gives sin θ 4 = 0.0419, so θ 4 = 2.40°.
We see from the diagram that
h1 ′ = h1 − R sin θ 3
= 1.0cm − (12.0cm ) sin1.60° = 0.665cm,
so the distance from the flat face to the point where the ray crosses the axis is
( 0.665cm ) = 15.9cm.
h′
d1 = 1 =
tan θ 4
tan 2.40°
(b) We find the incident angle from
( 4.0cm )
h
sin θ1 = 2 =
= 0.333, so θ1 = 19.47°.
R (12.0cm )
For the refraction at the curved surface, we have
sin θ1 = n sin θ 2 ;
sin19.47° = (1.50 ) sin θ 2 , which gives sin θ 2 = 0.222, so θ 2 = 12.84°.
We see from the diagram that
θ 3 = θ1 − θ 2 = 19.47° − 12.84° = 6.63°.
For the refraction at the flat face, we have
n sin θ 3 = sin θ 4 ;
(1.50 ) sin 6.63° = sin θ 4 ,
which gives sin θ 4 = 0.173, so θ 4 = 9.97°.
We see from the diagram that
h2 ′ = h2 − R sin θ 3 = 4.0cm − (12.0cm ) sin 6.63° = 2.61cm,
so the distance from the flat face to the point where the ray crosses the axis is
( 2.61cm ) = 14.8cm.
h′
d2 = 2 =
tan θ 4 tan 9.97°
(c) The separation of the “focal points” is
∆d = d1 − d 2 = 15.9cm − 14.8cm = 1.1cm.
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
245
Chapter 25
Optical Instruments
(d) When h2 = 4.0 cm, the rays focus closer to the lens, so they will form a circle at the “focal point”
for h1 = 1.0 cm. We find the radius of this circle from similar triangles:
h2 ′
r
=
;
d 2 ( d1 − d 2 )
( 2.61cm )
r
=
,
(14.8cm ) (1.1cm )
which gives r = 0.19cm.
48. The minimum angular resolution is
−9
1.22λ (1.22 ) ( 550 × 10 m )
=
= 2.64 × 10−7 rad = (1.51 × 10−5 ) °.
θ=
D
(100in )( 0.0254 m in )
49. The angular resolution of the eye, which is the required resolution using the telescope, is
−3
d eye ( 0.10 × 10 m )
=
= 4.0 × 10−4 rad.
θ eye =
−2
Leye
25
×
10
m
(
)
The resolution without the telescope is
( 7.0 km ) = 1.82 × 10−5 rad.
d
θ= =
L ( 3.84 × 105 km )
If we ignore the inverted image, the magnification is
θ eye f o
M=
= ;
θ
fe
( 4.0 × 10
(1.82 × 10
−4
−5
rad )
rad )
=
( 2.0 m ) ,
fe
which gives f e = 0.091m = 9.1cm.
The resolution limit is
−7
1.22λ (1.22 ) ( 5.5 × 10 m )
θ=
=
= 6.1 × 10−6 rad.
D
( 0.11m )
50. The minimum angular resolution is
1.22λ
.
θ=
D
The distance between lines is the resolving power:
1.22λ f
RP = f θ =
= 1.22λ ( f -stop ) .
D
f
we have
For
2
RP2 = (1.22 ) ( 550 × 10−9 m ) ( 2 ) = 1.34 × 10−6 m = 1.34 × 10−3 mm, so the resolution is
1
1
=
= 746lines mm.
RP2 (1.22 × 10−3 mm )
For
f
we have
16
RP2 = (1.22 ) ( 550 × 10−9 m ) (16.7 ) = 1.12 × 10−5 m = 1.12 × 10−2 mm, so the resolution is
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
246
Physics: Principles with Applications, 6th Edition
Giancoli
1
1
=
= 89lines mm.
RP2 (1.02 × 10−2 mm )
51. The resolution of the telescope is
−9
1.22λ (1.22 ) ( 550 × 10 m )
=
= 1.22 × 10−6 rad.
θ=
D
( 0.55m )
The separation of the stars is
d = Lθ = (15ly ) ( 9.46 × 1015 m ly )(1.22 × 10−6 rad ) = 1.7 × 1011 m.
52. (a) The resolution of the eye is
−9
1.22λ (1.22 ) ( 550 × 10 m )
θ=
=
= 1.34 × 10−4 rad.
−3
D
( 5.0 × 10 m )
We find the maximum distance from
d = Lθ ;
2.0 m = L (1.34 × 10−4 rad ) , which gives L = 1.5 × 104 m = 15km.
(b) The angular separation is the resolution:
θ = 1.34 × 10−4 rad = ( 7.68 × 10−3 ) ° = 0.46′.
Our answer is less than the real resolution because of the presence of aberrations.
53. The separation is s = 400 × 106 m. The distance away is L = 8 × 1010 m.
The angular separation is
s 400 × 106 m
= 0.005 rad.
θ= =
L
8 × 1010 m
The angular resolution of the eye is
−9
1.22λ (1.22 ) ( 550 × 10 m )
=
= 1.34 × 10−5 rad.
θe =
D
( 0.05 m )
Since θ > θ e , the answer is yes.
A person standing on Mars can resolve the Earth and its moon without
a telescope.
54. For the diffraction from the crystal, we have
mλ = 2d sin φ ; m = 1, 2,3,... .
For the first maximum, we get
(1)( 0.133nm ) = 2 ( 0.280 nm ) sin φ , which gives φ = 13.7°.
55. We find the spacing from
mλ = 2d sin φ ; m = 1, 2,3,... .
( 2 )( 0.0973nm ) = 2d sin 23.4°,
which gives d = 0.245nm.
56. (a) For the diffraction from the crystal, we have
mλ = 2d sin φ ; m = 1, 2,3,... .
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
247
Chapter 25
Optical Instruments
When we form the ratio for the two orders, we get
m2 ( sin φ2 )
;
=
m1 ( sin φ1 )
( sin φ2 )
2
, which gives φ2 = 58.4°.
=
1 ( sin 25.2° )
(b) We find the wavelength from
m1λ = 2d sin φ1 ;
(1) λ = 2 ( 0.24 nm ) sin 25.2°,
which gives λ = 0.20 nm.
57. (a) Because X-ray images are shadows, the image will be the
same size as the object, so the magnification is 1.
(b) The rays from the point source will not refract, so we can use
similar triangles to compare the image size to the object size
for the front of the body:
( d + d 2 ) (15cm + 25cm )
h
m1 = i = 1
=
= 2.7.
ho1
d1
(15cm )
For the back of the body, the image and object have the same
size, so the magnification is 1.
∗
ho1
d1
ho2
d2
58. (a) The focal length of the lens is
1
1
f = =
= 0.286 m = 28.6cm.
P 3.50D
(b) The lens produces a virtual image at his near point:
⎛ 1 ⎞ ⎛ 1 ⎞ 1
⎜
⎟ + ⎜ ⎟ = = P;
⎝ d o1 ⎠ ⎝ di1 ⎠ f
⎛ 1 ⎞ ⎛ 1 ⎞
⎜
⎟ + ⎜ ⎟ = + 3.50 D, which gives di1 = − 1.18m, so his near point is 120cm.
⎝ 0.23m ⎠ ⎝ di1 ⎠
(c) For Pam, we find the object distance that will have an image at her near point:
⎛ 1 ⎞ ⎛ 1 ⎞ 1
⎜
⎟+⎜
⎟ = = P;
⎝ d o2 ⎠ ⎝ di2 ⎠ f
⎛ 1 ⎞ ⎛
⎞
1
⎜
⎟+⎜
⎟ = + 3.50D, which gives d o2 = − 0.13m, which is 15 cm from her eyes.
⎝ d o2 ⎠ ⎝ − 0.23m ⎠
59. The exposure is proportional to the intensity, the area and the time:
Exposure ∝ IAt ∝ ID 2 t ∝ It
.
2
( f -stop )
With the same shutter speed, the time is constant.
Because we want the exposure to be the same, we have
I1
I2
=
;
2
2
( f -stop1 ) ( f -stop 2 )
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
248
hi
Physics: Principles with Applications, 6th Edition
Giancoli
I1
( 5.6 )
2
=
I2
( 22 )
2
, which gives I 2 = 16 I1 .
Note that we have followed convention to use multiples of 2.
60. When an object is very far away, the image will be at the focal point di = f . Thus the magnification is
d
f
m = − i = − , that is, proportional to f.
do
do
61. For the magnification, we have
h
d
m = − i = − i = −1, so di = d o .
ho
do
We find the object distance from
⎛ 1 ⎞ ⎛1⎞ 1
⎜ ⎟ +⎜ ⎟= ;
⎝ d o ⎠ ⎝ di ⎠ f
⎛ 1 ⎞ ⎛ 1 ⎞
1
, which gives d o = 100 mm.
⎜ ⎟ +⎜ ⎟=
⎝ d o ⎠ ⎝ d o ⎠ 50 mm
The distance between the object and the film is
d = d o + d i = 100 mm + 100 mm = 200 mm.
62. The actual far point of the person is 155cm. With the lens, an object far away is to produce a virtual
image 155cm from the eye, or 153cm from the lens.
We find the power of the lens from
⎛ 1 ⎞ ⎛ 1 ⎞ 1
⎜
⎟ + ⎜ ⎟ = = P1 , when distances are in m;
⎝ d o1 ⎠ ⎝ d i1 ⎠ f1
⎞
1
⎛1⎞ ⎛
⎟ = P1 = − 0.65D (upper part).
⎜ ⎟+⎜
⎝ ∞ ⎠ ⎝ − 1.53m ⎠
The actual near point of the person is 45cm. With the lens, an object placed at the normal near point,
25 cm, or 23 cm from the lens, is to produce a virtual image 45cm from the eye, or 43cm from the lens.
We find the power of the lens from
⎛ 1 ⎞ ⎛ 1 ⎞ 1
= P2 ;
⎜
⎟+⎜
⎟=
⎝ d o2 ⎠ ⎝ di2 ⎠ f 2
⎛ 1 ⎞ ⎛
⎞
1
⎜
⎟+⎜
⎟ = P2 = + 2.0 D (lower part).
⎝ 0.23m ⎠ ⎝ − 0.43m ⎠
63. The maximum magnification is achieved with the image at the near point:
(15.0cm )
N
M1 = 1 + 1 = 1 +
= 2.9 × .
f
( 8.0cm )
For an adult we have
( 25.0cm )
N
M2 =1 + 2 =1 +
= 4.1 × .
f
( 8.0cm )
The person with the normal eye (adult) sees more detail.
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
249
Chapter 25
Optical Instruments
64. The magnification for a relaxed eye is
N
M = = NP = ( 0.25m )( + 4.0 D ) = 1.0 × .
f
65. (a) The magnification of the telescope is given by
( 4.5D )
f
P
M =− o =− e =−
= − 2.25 × .
fe
Po
( 2.0 D )
(b) To get a magnification greater than 1, for the eyepiece we use the lens with the smaller focal
length, or greater power: 4.5 D.
66. To find the new near point, we have
⎛ 1 ⎞ ⎛ 1 ⎞ 1
⎜
⎟ + ⎜ ⎟ = = P1 , when distances are in m;
⎝ d o1 ⎠ ⎝ d i1 ⎠ f1
⎛ 1 ⎞ ⎛ 1 ⎞
⎜
⎟ + ⎜ ⎟ = + 2.5D, which gives di1 = − 2.8m.
⎝ 0.35m ⎠ ⎝ di1 ⎠
To give him a normal near point, we have
⎛ 1 ⎞ ⎛ 1 ⎞ 1
= P2 ;
⎜
⎟+⎜
⎟=
⎝ d o2 ⎠ ⎝ di2 ⎠ f 2
⎛ 1 ⎞ ⎛ 1 ⎞
⎜
⎟+⎜
⎟ = P2 = + 3.6 D.
⎝ 0.25m ⎠ ⎝ − 2.8m ⎠
67. For the minimum aperture the angle subtended at the lens by the smallest feature is the angular
resolution:
d 1.22λ
θ= =
;
L
D
( 5 × 10−2 m ) = (1.22 ) ( 550 × 10−9 m ) , which gives D = 0.34 m = 34cm.
D
( 25 × 103 m )
68. (a) When an object is very far away, the image will be at the focal point di1 = f .
From the magnification, we have
h
d
m = − i = − i , so we see that hi is proportional to di .
ho
do
When the object and image are the same size, we get
h
d
− i = −1 = − i2 , so d o2 = di2 .
ho
d o2
From the lens equation, we get
⎛ 1 ⎞ ⎛ 1 ⎞ 1
⎜
⎟+⎜
⎟= ;
⎝ d o2 ⎠ ⎝ di2 ⎠ f
⎛ 1 ⎞ ⎛ 1 ⎞ 1
⎜
⎟+⎜
⎟ = , which gives di2 = 2 f .
⎝ d i2 ⎠ ⎝ di2 ⎠ f
The required exposure time is proportional to the area of the image on the film:
2
2
t ∝ A ∝ ( hi ) ∝ ( di ) .
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
250
Physics: Principles with Applications, 6th Edition
Giancoli
When we form the ratio, we get
2
2
t2 ⎛ di2 ⎞ ⎛ 2 f ⎞
=⎜
⎟ =⎜
⎟ = 4.
t1 ⎝ di1 ⎠ ⎝ f ⎠
(b) For the increased object distances, we have
⎛ 1 ⎞ ⎛ 1 ⎞ 1
⎜
⎟ +⎜ ⎟= ;
⎝ d o3 ⎠ ⎝ di3 ⎠ f
⎛ 1 ⎞ ⎛ 1 ⎞
⎜
⎟ +⎜ ⎟=
⎝ 4 f ⎠ ⎝ di3 ⎠
⎛ 1 ⎞ ⎛ 1 ⎞
⎜
⎟+⎜
⎟=
⎝ d o4 ⎠ ⎝ di4 ⎠
2
t ⎛4⎞
4f
1
; and 3 = ⎜ ⎟ = 1.78.
, which gives d i3 =
3
t1 ⎝ 3 ⎠
f
1
;
f
2
⎛ 1 ⎞ ⎛ 1 ⎞ 1
t ⎛5⎞
5f
; and 4 = ⎜ ⎟ = 1.56.
⎟ = , which gives di4 =
⎜
⎟+⎜
4
t1 ⎝ 4 ⎠
⎝ 5 f ⎠ ⎝ d i4 ⎠ f
These increased exposures are less than the minimal adjustment on a typical camera, so they are
negligible.
69. The focal length of the eyepiece is
1
1
= 4.3 × 10−2 m = 4.3cm.
fe = =
Pe 23D
For both object and image far away, we find the focal length of the objective from the separation of the
lenses:
L = fo + fe ;
85cm = f o + 4.3cm, which gives f o = 80.7 cm.
The magnification of the telescope is given by
( 80.7 cm )
f
M =− o =−
= −19 × .
fe
( 4.3cm )
70. The distance from the Earth to the Moon is 384 × 106 m.
s = Lθ =
L (1.22λ )
D
( 384 × 10 m ) (1.22 ) ( 550 × 10
=
6
( 2.4 m )
−9
m)
= 107 m.
71. (a) The distance between the two lenses is f o + f e = 4.0cm + 44cm = 48cm.
The magnification is
( 44cm )
f
M =− o =−
= −11 × .
fe
( 4.0cm )
The eyepiece is the lens with f = 4.0cm.
(b) The magnification is
( 25 cm ) l
Nl
M =−
=
= 25.
f e f o ( 44 cm )( 4.0 cm )
Solve for l to get l = 180cm.
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
251
Chapter 25
Optical Instruments
fo
= − 8.0. Therefore, f o = 8.0 f e .
fe
Since f o + f e = 28cm, 8.0 f e + f e = 9.0 f e = 28cm.
28cm
= 3.1cm and f e = 24.9cm.
Solving, f e =
9.0
72. M = −
73. The angular resolution must equal
( 0.05 m )
s
θ= =
= 3.85 × 10−7 .
L (130 × 103 m )
Therefore,
−9
1.22 λ (1.22 ) ( 550 × 10 m )
=
= 3.8 × 10−7 .
D
D
Solve for D to get D = 1.7 m.
θ=
74. We find the focal lengths of the lens for the two colors:
⎛ 1
1
1 ⎞
= ( nred − 1) ⎜ +
⎟
f red
⎝ R1 R2 ⎠
1
f orange
⎡⎛ 1 ⎞ ⎛ 1 ⎞ ⎤
= (1.5106 − 1) ⎢⎜
⎟ + ⎜ ⎟ ⎥ which gives f red = 36.04cm.
⎣⎝ 18.4cm ⎠ ⎝ ∞ ⎠ ⎦
⎛ 1
1 ⎞
= ( norange − 1) ⎜ + ⎟
⎝ R1 R2 ⎠
⎡⎛ 1 ⎞ ⎛ 1 ⎞ ⎤
= (1.5226 − 1) ⎢⎜
⎟ + ⎜ ⎟ ⎥ which gives f orange = 35.21cm.
⎣⎝ 18.4cm ⎠ ⎝ ∞ ⎠ ⎦
We find the image distances from
⎛ 1 ⎞ ⎛ 1 ⎞ 1
;
⎜ ⎟+⎜
⎟=
⎝ d o ⎠ ⎝ d ired ⎠ f red
⎛ 1 ⎞ ⎛ 1 ⎞
1
cm, which gives d ired = 79.4cm.
⎟=
⎜
⎟+⎜
⎝ 66.0cm ⎠ ⎝ d ired ⎠ 36.04
⎛ 1 ⎞ ⎛ 1 ⎞
1
;
⎟⎟ =
⎜ ⎟ + ⎜⎜
⎝ d o ⎠ ⎝ d iorange ⎠ f orange
⎛ 1 ⎞ ⎛ 1 ⎞
1
, which gives diorange = 75.5cm.
⎟⎟ =
⎜
⎟ + ⎜⎜
⎝ 66.0cm ⎠ ⎝ diorange ⎠ 35.21cm
The images are 3.9 cm apart, an example of chromatic aberration.
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
252