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CHAPTER 24
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
Geometrical Optics & ...
Answers to Discussion Questions
•24.1
The focal length increases because the rays are not bent as strongly at the water-glass interface.
•24.2
It will be positive. It’s a diverging lens.
•24.3
The focal length depends on the index of refraction and that depends on the wavelength.
•24.4
They provide an air space in front of the cornea so that the eye can work with its normal amount
of refraction at that first interface.
•24.5
Form a real image of a very distant object; the image-distance then approaches the focal length.
Place the two in contact, shine in parallel light and measure f , knowing that 1/f = 1/f+ +1/f– ,
where f+ is given and f– is to be found.
•24.6
The rump in the real, the head in the virtual.
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•24.7
The gas provides a medium with an index greater than one and so lessens the bending of the
rays. As it is pumped out, the rays are brought to convergence more strongly and the focal
point moves toward the lens.
•24.8
The rays must have been converging to begin with.
•24.9
The radius of curvature is ∞ and so is f . That means the object- and image-distances must
have equal magnitudes. Thus, the magnification is +1.
•24.10
A point source emits spherical waves which strike the surface and appear to come from a point
source at the other focus. Hence the reflected waves are spherical too.
•24.11
If the soldiers lined up along a parabolic arc with the ship’s sails at the focus of the parabola,
parallel rays from the Sun reflected off their shields would converge on the ship.
•24.12
The focal length is shorter and so the dioptric power is greater.
•24.13
The target is at one of the two focii of the hyperboloid and rays reflected from it appear to
come from the other focus, F1 (H), but this is also a focus, F1 (E), of the ellipsoidal mirror.
Rays appearing to come from one focus of the ellipsoid, after reflecting off it, converge to the
other focus, F2 (E), at the film plane.
•24.14
The rays will be more converging than necessary so objects at infinity (and far away) will
be blurry because the eye cannot handle converging rays. Nearer objects with more strongly
diverging rays will be seen clearly.
•24.15
The object has a diameter d, where d(1200) = 0.0005 m, d = 4.2 × 10−7 m, which is the same
as the wavelength of violet light. We cannot hope to see objects that are smaller than the
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probe being used; namely, the wavelength of light. The amount of diffraction will then obscure
the image completely. So the details we wish to observe cannot be finer than λ, which puts a
practical limit on the magnification. A 12000× microscope will only make larger blurred images
showing no more detail.
•24.16
The special flat filament of the lamp is at the center of the curvature of the spherical mirror so
as to reflect light back toward the filament which is at the focus of the lens. Thus most of the
light emerges as a parallel beam. By shifting the location of the lamp with respect to the lens,
the shape of the beam can be changed.
•24.17
The filament is located at one focus of the ellipsoid and the reflected light is made to converge
at the other focus, which also corresponds to the front focal point of the lens combination.
•24.18
geometrical optics —– An idealized domain of optics, where the diffraction of light is negligible and all the light travel in straight lines in accordance with the Laws of Reflection and
Refraction.
lens —– A ground or molded piece of glass, plastic, or other transparent material with opposite
surfaces either or both of which are curved, by means of which light rays are refracted so that
they converge or diverge to form an image.
aspherical surfaces —– Surfaces whose shape do not resemble part of a sphere.
converging lens —– A lens which causes the incoming rays to converge to some extent, to
bend towards the central axis.
diverging lens —– A lens which turns the incoming rays outward away from the central axis.
concave —– Curved like the inner surface of a sphere.
convex —– Having a surface or boundary that curves or bulges outward, as the exterior of a
sphere.
paraxial —– Rays that are not far from the central axis and enter only at shallow angles are
known to be paraxial.
thin lens —– Lenses for which the radii of curvature of the surfaces are large compared to
the thickness.
object distance —– The distance between the object and the optical instrument (such as a
lens or a mirror).
image distance —– The distance between the optical instrument (such as a lens or a mirror)
and the image it forms.
Thin-Lens Equation —– 1/so + 1/si = (nl − 1)(1/R1 − 1/R2 ).
Lensmaker’s Formula —– 1/f = (nl − 1)(1/R1 − 1/R2 ).
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focal point —– A point at which rays of light or other radiation converge or from which they
appear to diverge, as after refraction or reflection in an optical system.
focal plane —– The plane containing the focal point, on which parallel rays focus.
positive lens —– Converging lens.
negative lens —– Diverging lens.
optical center —– A special point at the center of any thin lens, such that any paraxial ray
heading toward it will pass through undeviated and may be drawn as a straight line.
real image —– An image formed by converging light rays.
virtual image —– An image from which rays of reflected or refracted light appear to diverge.
transverse magnification —– The ratio of any transverse (i.e., perpendicular to the optical
axis) dimension of the image formed by an optical system to the corresponding dimension of
the object.
longitudinal magnification —– The ratio of any longitudinal (i.e., parallel to the optical
axis) dimension of the image formed by an optical system to the corresponding dimension of
the object.
cornea —– The transparent convex anterior portion of the outer fibrous coat of the eyeball
that covers the iris and the pupil and is continuous with the sclera.
crystalline lens —– A transparent, biconvex body of the eye between the iris and the vitreous
humor that focuses light rays entering through the pupil to form an image on the retina.
retina —– A delicate, multilayered, light-sensitive membrane lining the inner eyeball and
connected by the optic nerve to the brain.
accommodation —– The fine focusing performed by the human eye.
far-point —– The point that is seen clearly by the unaccomodated eye.
near-point —– The closest point that can be clearly seen with maximum accommodation.
angular magnification —– The ratio of the size of the retinal image formed by the device
to the size of the retinal image formed by the unaided eye at normal viewing distance.
magnifying glass —– A lens or combination of lenses that enlarges the image of an object.
It is usually a convex lens placed within one focal length of the object.
intermediate image —– An image formed by one of a series of optical instrument and one
which serves as the effective object of the next instrument.
dioptic power —– A measure of the focusing power of a lens, defined as the reciprocal of
the focal length.
farsightedness —– An abnormal condition of the eye in which vision is better for distant
objects than for near objects. It results from the eyeball being too short from front to back,
causing images to be focused behind the retina.
nearsightedness —– A visual defect in which distant objects appear blurred because their
images are focused in front of the retina rather than on it
objective eyepiece —– The lens or lens group closest to the eye in an optical instrument.
mirror formula —– 1/so + 1/si = −2/R.
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Answers to Multiple Choice Questions
1. b
8. b
15. c
22. a
2. c
9. a
16. e
23. d
3. b
10. c
17. e
24. c
4. c
11. b
18. b
25. b
5. d
12. d
19. a
6. d
13. c
20. d
7. b
14. e
21. d
Solutions to Problems
24.1
First, use the Lensmaker’s Formula, Eq. (24.2), to find f :
−1 −1
1
1
1
1
f = (nl − 1)
−
= (1.50 − 1)
−
= 0.50 m .
R1
R2
0.50 m −0.50 m
Now locate the image from Eq. (24.4), 1/so + 1/si = 1/f , or
si =
so f
(1.00 m)(0.50 m)
= 1.0 m ,
=
so − f
1.00 m − 0.50 m
i.e., the image is located 1.0 m to the right of the lens.
24.2
With the previous problem in mind, we first find f :
−1 −1
1
1
1
1
f = (nl − 1)
−
= (1.50 − 1)
−
= 1.5 m .
R1
R2
1.00 m −3.00 m
Now find so from Eq. (24.4), 1/so + 1/si = 1/f :
so =
si f
(2.00 m)(1.5 m)
= 6.0 m ,
=
si − f
2.00 m − 1.5 m
i.e., the light source should be positioned 6.0 m in front of the lens.
24.3
Let R1 = −R2 = R for the lens. Then Eq. (24.1) gives 1/so + 1/si = (nl − 1)(1/R + 1/R) =
2(nl − 1)/R. Plug in so = si = 24.0 cm, and nl = 1.50, and solve for R:
R=
2(nl − 1)
2(1.50 − 1)
=
= 12 cm .
1/so + 1/si
1/(24.0 cm) + 1/(24.0 cm)
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24.4
The lens is convex since it is thicker in the middle. The focal length f is given by Eq. (24.2),
with nl = 1.58, R1 = 1.00 m, and R2 = ∞ (for the flat surface):
−1 −1
1
1
1
1
f = (nl − 1)
−
= (1.50 − 1)
−
= 2.0 m .
R1
R2
1.00 m ∞
So the sunlight (with so ≈ ∞) will be focused a distance si = f = 2.0 m behind the lens.
24.5
Use Eq. (24.2) for the focal length f , with nl = 1.58, R1 = +1.00 m, and R2 = −1.00 m:
−1 −1
1
1
1
1
f = (nl − 1)
−
= (1.58 − 1)
−
= 0.86 m .
R1
R2
1.00 m −1.00 m
24.6
For a bi-convex lens one of the two radii of curvature is negative. Since |R1 | = 2.0 m < |R2 | =
5.0 m we must take R1 = 2.0 m and R2 = −5.0 m to ensure that f > 0. Plug these into
Eq. (24.2), along with nl = 1.5, to find f :
−1 −1
1
1
1
1
f = (nl − 1)
−
= (1.5 − 1)
−
= 2.9 m .
R1
R2
2.0 m −5.0 m
24.7
From the Lensmaker’s Formula [Eq. (24.2)]
−1 −1
1
1
1
1
f = (nl − 1)
−
= (1.50 − 1)
−
= −1.0 m .
R1
R2
0.500 m 0.250 m
Note that f < 0 since it’s a concave lens.
24.8
Similar to the previous problem, this time R1 = 0.250 m and R2 = 0.500 m, so
−1 −1
1
1
1
1
f = (nl − 1)
−
= (1.50 − 1)
−
= +1.0 m .
R1
R2
0.250 m 0.500 m
Note that f > 0 since it’s a convex lens.
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24.9
The focal length f of the lens is given by the Lensmaker’s formula, 1/f = (nl −1)(1/R1 −1/R2 ).
In this case f = 1.60 m, R1 = 2.00 m, R2 = ∞ (for the planar surface), which we plug into the
formula to solve for nl :
nl = 1 +
1
1
=1+
= 2.25 .
f (1/R1 − 1/R2 )
(1.60 m) (1/2.00 m − 1/∞)
24.10
It is clear that f = 2.50 m for the lens, since si = 2.50 m when so = ∞. Now set so = 4.00 m
and find si from Eq. (24.4), 1/so + 1/si = 1/f :
si =
(4.00 m)(2.50 m)
so f
=
= 6.67 m ,
so − f
4.00 m − 2.50 m
i.e., the screen must be positioned at 6.67 m behind the lens. Since si > 0 the image is real.
24.11
From the problem statement we know that si = ∞ when so = 25.0 cm. Plug these into
Eq. (24.4) to obtain f = si = 25.0 cm. Now set so = 150 cm and find the corresponding value
of si from Eq. (24.4):
si =
so f
(150 cm)(25.0 cm)
= 30.0 cm ,
=
so − f
150 cm − 25.0 cm
i.e., the image is at 30.0 cm behind the lens. Again, since si > 0 the image is real.
24.12
For a plane wave (with parallel light rays) so = ∞, so f = si = 100 cm. Now replace the value
of so with 200 cm and find the new value of si from Eq. (24.4):
si =
so f
(200 cm)(100 cm)
= 200 cm .
=
so − f
200 cm − 100 cm
Again, since si > 0 the image is real.
24.13
Use the Lensmaker’s Formula: 1/f = (nl − 1)(1/R1 − 1/R2 ). In this case f , R, and R2 are
all positive, so R2 > R1 > 0. Set f = 100 cm, nl = 3/2, and R2 = 100 cm and solve for R1 ,
the smaller of the two radii:
−1 −1
1
1
1
1
R1 =
=
= 33.3 cm .
+
+
f (nl − 1) R2
(100 cm)(3/2 − 1) 100 cm
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24.14
Apply Eq. (24.1): 1/so + 1/si = (nl − 1)(1/R1 − 1/R2 ). In this case R2 = ∞, n = 1.50,
so = 50.0 cm, and si = ∞. Plug these date into the equation above and solve for R1 :
R1 =
1/so + 1/si
1
+
nl − 1
R2
−1
=
1/50.0 cm + 1/∞
1
+
1.50 − 1
∞
−1
= 25.0 cm .
Thus the radii of the lens are 25.0 cm and ∞.
24.15
In deriving Eq. (24.2), 1/fa = (nl − 1)(1/R1 − 1/R2 ), we made use of the Law of Refraction
for a light ray entering the lens from air: ni sin θi = nt sin θt , or sin θt = (nt /ni ) sin θi . Here i
stands for air and t for the glass. Taking ni = nair = 1 and nt = nl , this becomes
sin θt = nl sin θi .
In the current case i stands for water (w), so
sin θt =
nt
ni
sin θi =
nl
nw
sin θi .
Comparing this expression with that in the previous case, we find that, to obtain the new focal
length fw , we must replace nl with nl /nw when the lens is submerged in water in stead of air.
Thus Eq. (24.2) becomes
1
nl
1
1
.
=
−1
−
fw
nw
R1
R2
Numerically, the factor on the RHS of the equation is now nl /nw − 1 = 1.5/(4/3) − 1 = 0.125,
which is to be compared with the old factor of nl − 1 = 0.5 when the lens is submerged in air.
Thus
fa
0.125
1
n /n − 1
=
= ,
= l w
fw
nl − 1
0.5
4
or fw = 4fa .
24.16
(a) so is the distance between the lens and the Sun, and is virtually infinity.
(b) so > 2f .
(c) Real, since so > f .
(d) Inverted, which is the case for all real images formed by a positive lens.
(e) minified, since so > 2f .
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(f)
f
f
(g) Use Eq. (24.4), the Gaussian Lens Equation, with so = ∞ and f = +80.0 cm, to find si :
1/f = 1/so + 1/si , or 1/si = 1/f + 1/so = 1/f + 1/∞ = 1/f , and so si = f = 80.0 cm.
24.17
(a) so = 3.00 m (the distance between the object and the lens).
(b) so = 3.00 m and f = 100.0 cm = 1.000 m, so so > 2f .
(c) Since so > f the image is real and can be projected onto a screen.
(d) Inverted, which is the case for all real images formed by a positive (convex) lens.
(e) minified, since so > 2f .
(f)
O
f
f
I
(g) Use Eq. (24.4), the Gaussian Lens Equation, with so = 3.00 m and f = 1.000 m, to find si :
1/f = 1/so + 1/si ,
si =
(3.00 m)(1.000 m)
so f
=
= +1.50 m ,
so − f
3.00 m − 1.000 m
i.e., the image of the person is located 1.50 m behind the lens. It is real (since si > 0), with a
magnification of
s
1.50 m
MT = − i = −
= −0.500 ,
so
3.00 m
i.e., it is half-sized and inverted (since MT < 0). Here we used Eq. (24.7) for the transverse
magnification.
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813
24.18
The object distance (300 cm) is much greater than the focal length (5.00 cm), and much greater
than 2f , hence the image is real, inverted, minified, and located near the focal plane (si ≈ f ).
Now use Eq. (24.4), with so = 300 cm and f = +5.00 cm, to find si : 1/f = 1/so + 1/si , or
si =
(300 cm)(5.00 cm)
so f
=
= +5.1 cm ,
so − f
300 cm − 5.00 cm
i.e., the image is located 5.1 cm behind the lens, with a magnification of
MT = −
si
5.1 cm
=−
= −0.017 ,
so
300 cm
meaning that the image is much less in size than the object.
24.19
(a) so = 12.0 cm (the distance between the object and the lens).
(b) si = −4.8 cm, which is the distance between the image and the lens — it is negative since
it is to the left of the lens, the same side as the object.
(c) That would be Eq. (24.4), the Gaussian Lens Equation: 1/f = 1/so + 1/si .
(d)
1
s s
(12.0 cm)(−4.8 cm)
f=
= o i =
= −8.0 cm .
1/so + 1/si
so + si
12.0 cm − 4.8 cm
(e) f < 0 means that it is a negative (concave) lens.
(f) The image is virtual and therefore cannot be projected onto a screen.
(g) See Fig. 24.17 in the text.
(h) Upright, as is the case for any image formed by a concave lens.
(i) See Fig. 24.17 in the text.
24.20
Use Eq. (24.4), the Gaussian Lens Equation, with so = 30.0 cm and f = +10.0 cm, to find si :
1/f = 1/so + 1/si , or
si =
so f
(30.0 cm)(10.0 cm)
= +15.0 cm ,
=
so − f
30.0 cm − 10.0 cm
i.e., the image of the cat is located 15.0 cm behind the lens. It is real (since si > 0), with a
magnification of
s
15.0 cm
MT = − i = −
= −0.500 ,
so
30.0 cm
i.e., it is half-sized and inverted (since MT < 0). Here we used Eq. (24.7) for the transverse
magnification.
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24.21
With the previous problem in mind, we use Eq. (24.4), with so = 100.0 cm and f = 60.0 cm,
to find si :
s f
(100.0 cm)(60.0 cm)
si = o
= +150 cm ,
=
so − f
100.0 cm − 60.0 cm
i.e., the image of the frog is located 150 cm behind the lens. It is real (since si > 0), with a
magnification of
s
150 cm
MT = − i = −
= −1.50 ,
so
100.0 cm
so the image is 50% larger than the frog itself and inverted (since MT < 0).
24.22
Similar to the previous two problems, we again apply Eq. (24.4), 1/so + 1/si = 1/f , with
so = 100 cm and f = 50.0 cm, to find si :
si =
(100 cm)(50.0 cm)
so f
=
= +100 cm ,
so − f
100 cm − 50.0 cm
i.e., the image of the flower is located 100 cm behind the lens. It is real (since si > 0), with a
magnification of
s
100 cm
MT = − i = −
= −1.00 ,
so
100 cm
so the image is life-sized (since |MT | = 1.00) and inverted (since MT < 0).
24.23
The location of the image is a distance si from the lens, where si is (see the previous three
problems)
s f
(80.0 mm)(120 mm)
si = o
= −240 mm ,
=
so − f
80.0 mm − 120 mm
where we plugged in so = 80.0 mm and f = 120 mm. Note that here si < 0, so image is virtual
and a distance 240 mm in front of the lens (i.e., on the same side of the lens as the object).
24.24
The image produced by a concave lens is always virtual, minified, and right-side-up. To locate
the image of the beacon light, plug in so = 50 m and f = −10 m into Eq. (24.4) and solve for
si :
s f
(50 m)(−10 m)
si = o
=
= −8.3 m ,
so − f
50 m − (−10 m)
so the image appears to be located 8.3 m outside the window (on the same side as the object).
You can easily verify that 0 < MT = −si /so < 1, meaning that the image is minified and
right-side-up.
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24.25
(a) First, find the location of the image by solving for si from Eq. (24.4), which gives
si =
so f
(0.75 m)(0.25 m)
= 0.38 m ,
=
so − f
0.75 m − 0.25 m
so the image is located a distance 0.38 m behind the lens. It is real (since si > 0), with a
magnification of MT = −si /so = −(0.375 m)/(0.75 cm) = −0.50, so the image is half-sized
(since |MT | = 0.50) and inverted (since MT < 0).
(b) The ray diagram is shown below.
f
f
24.26
The eyeglass is a positive (convex) lens with f = +2.0 m. The location of the image of the bus is
at 2.5 m behind the glass, so si = +2.5 m. Now solve for so from Eq. (24.4), 1/f = 1/so + 1/si :
so =
si f
(2.5 m)(2.0 m)
= 10 m ,
=
si − f
2.5 m − 2.0 m
i.e., the bus is a distance 10 m away from the eyeglass.
24.27
The lens is positive (convex) since it has to focus a real image onto the film, so f = +60.0 mm.
Also, to focus the image of the bug on the film the distance between the lens and the film should
be si , so si = 100 mm. It follows that (see the previous problem)
so =
si f
(100 mm)(60.0 mm)
= 150 mm ,
=
si − f
100 mm − 60 mm
so the bug should be a distance 150 mm away from the camera lens.
24.28
With the previous problem in mind, we plug f = +52.0 mm, along with so = 5.00 m, into
Eq. (24.4) and solve for si , the desired distance between the lens and the film:
si =
so f
(5.00 m)(52.0 × 10−3 m)
= 0.052 5 m = 52.5 mm .
=
so − f
5.00 m − 52.0 × 10−3 m
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Since the camera is originally focused on a distance object (i.e., so ≈ ∞) the original distance
between the lens and the film is the same as the focal length, at 52.0 mm. So to re-focus when
the object is 5.00 m from the lens the lens has to be advanced by 52.5 mm − 52.0 mm = 0.5 mm.
24.29
When the eye is in the relaxed state we may use Eq. (24.9) to find the magnification of the lens
to be MA = dn /f , where dn = 0.254 m is the normal viewing distance (to the near-point) for
a standard observer. In this case MA = 2.0×, so the focal length of the lens is
f=
dn
0.254 m
=
= 0.13 m .
MA
2.0
24.30
Similar to the previous problem, we apply Eq.(24.9) for the magnification MA , with dn = 25 cm
and f = 10 cm:
d
25 cm
MA = n =
= 2.5× .
f
10 cm
24.31
The focal length f of the magnifier is related to its magnification MA via MA = dn /f , where
dn = 0.254 m for a standard observer. Since MA = 8×,
f=
dn
0.254 m
=
= 0.03 m = 3 cm ,
MA
8
so the lens should be placed 3 cm above the flat surface.
24.32
(a) so = 10.0 cm (the distance between the object and the lens). while f = 20.0 cm, so so < f .
(b) Virtual, since so < f .
(c) Erect, since so < f .
(d) Magnified, since so < f .
(e)
I
O
f
f
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(f) Apply Eq. (24.4), 1/f = 1/so + 1/si :
si =
so f
(10.0 cm)(20.0 cm)
= −20.0 cm ,
=
so − f
10.0 cm − 20.0 cm
i.e., the image of the candle is located 20.0 cm to the left of the lens.
(g)
MT = −
si
−20.0 cm
=−
= +2.00 .
so
10.0 cm
(h) MT > 0 means that the image is erect.
(i) Since MT = 2.00 the image is twice the size of the object, or (2.00)(1.0 cm) = 2.0 cm tall.
24.33
(a) so = 12.0 cm (the distance between the object and the lens). si = −4.0 cm, which is the
distance between the image and the lens — it is negative since it is on the same side as the
object.
(b) Use Eq. (24.4), the Gaussian Lens Equation: 1/f = 1/so + 1/si , to obtain
f=
1
s s
(12.0 cm)(−4.0 cm)
= o i =
= −6.0 cm .
1/so + 1/si
so + si
12.0 cm − 4.0 cm
(c) f < 0 means that it is a negative (concave) lens.
(d) The image is virtual and therefore cannot be projected onto a screen.
(e)
f
I
f
O
(f) Upright, as is the case for any image formed by a concave lens.
(g)
MT = −
si
−4.0 cm
=−
= +0.33 .
so
12.0 cm
(h) MT > 0 means that the image is erect.
(i) Since MT = 0.333 the image of the butterfly is 0.333 times its original size, or (0.333)(3.0 cm)
= 1.0 cm.
818
CHAPTER 24
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
24.34
First, compute the focal length f of the lens from Eq. (24.2):
−1 −1
1
1
1
1
f = (nl − 1)
−
= (1.5 − 1)
−
= 0.60 m .
R1
R2
0.60 m −0.60 m
(a) For a life-sized, real image MT = −si /so = −1.0, so so = si , and Eq. (24.4) now becomes
1/f = 1/so + 1/so = 2/so , which gives
so = 2f = 2(0.60 m) = 1.2 m ,
i.e., the object is to be placed 1.2 m in front of the lens. Here we noted that the image must be
real since it can be projected onto a screen, and so si > 0.
(b) Since si = so (see the previous part), si = 1.2 m as well, meaning that the screen is to be
placed 1.2 m behind the lens for a sharp image to be focused on it.
(c)
2f
f
2f
f
24.35
When an object is placed a distance so in front of a convex lens of focal length f a real image is
formed a distance si behind the lens (if so > f ). Here so , si and f are related via the Gaussian
Lens Equation [Eq. (24.4)]: 1/f = 1/so + 1/si . In our case the object (grasshopper) is located
10 cm to the left of the lens while its image is 30 cm to the right of the lens, so so = 10 cm and
si = 30 cm. Plug these data into Eq. (24.4) to find f , the focal length of the lens:
1
1
1
1
4
1
+
=
+
=
,
=
f
so
si
10 cm 30 cm
30 cm
and so f = 30 cm/4 = 7.5 cm.
(a) Now the grasshopper jumps 7.5 cm towards the lens so so decreases by 7.5 cm, to 10 cm −
7.5 cm = 2.5 cm. Plug this new value of so , along with f = 7.5 cm, into Eq. (24.4) again and
solve for si , the new location of the image:
si =
so f
(2.5 cm)(7.5 cm)
= −3.8 cm ,
=
so − f
2.5 cm − 7.5 cm
CHAPTER 24
819
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
which means that the image is now 3.8 cm to the left of the lens (i.e., on the same side of the
lens as the grasshopper itself), since si < 0.
(b) At first si = 30 cm and so = 10 cm, so the transverse magnification is MT = −si /so =
−30 cm/10 cm = −3.0. The image is real (as si > 0), inverted (as MT < 0), and magnified to
three times the original size of the grasshopper (as |MT | = 3.0).
Similarly, for the new image si = −3.8 cm < 0, so it is now virtual. The transverse magnification
is now MT = −si /so = −3.8 cm/2.5 cm = +1.5 > 0, so the new image is right-side-up and is
magnified to 1.5 times the original size.
(c) The two ray diagrams below depict the situation before and after the grasshopper makes
the jump.
f
f
f
f
24.36
For an image to be projected onto a screen it must be real, so si = +10 m. Also, it is enlarged
100 times so |MT | = |si /so | = 100, which gives so = si /100 = 10 m/100 = 0.10 m. Plug the
values of so and si into Eq. (24.4) to find the desired focal length f of the projector lens:
f=
1
1
+
so
si
−1
=
1
1
+
0.10 m 10 m
−1
= 0.099 m = 9.9 cm .
24.37
The size of Fred-the-chicken’s face is yo = 5.0 cm, while the size of the corresponding image on
the film is yo = 24 mm. Thus the transverse magnification MT provided by the camera lens is
820
CHAPTER 24
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
[see Eqs (24.8) and (24.9)] MT = −si /so = yi /yo = 0.24 cm/5.0 cm. Now plug in so = 2.0 m
into this equation to find si = 0.96 m. The focal length f of the camera then follows from
Eq. (24.4):
f=
1
1
+
so
si
−1
=
1
1
+
2.0 m 0.96 m
−1
= 0.65 m = 6.5 × 102 mm .
24.38
Since the Moon is very far away from us we may take so = 3.84 × 108 m ≈ ∞, which yields
si ≈ f = 50 mm by virtue of Eq. (24.4). The diameter of the Moon is yo = 0.273(1.27 ×
106 m) = 3.467 × 106 m. Plug these values into Eqs. (24.8) and (24.9), MT = −si /so = yi /yo ,
to find the size (diameter) of the image of the Moon on the film:
si yo (50 mm)(3.467 × 106 m)
=
|yi | = −
= 0.45 mm .
si 3.84 × 108 m
(If you left out the absolute value sign you would get yi = −0.45 mm, where the minus sign
only means that the image of the Moon is inverted.)
24.39
Consider the two similar triangles in Fig. P39, one with a base yo and corresponding height so ,
and the other with a base yi and corresponding height si . By similarity |yo /so | = |yi /si |, or
|yi /yo | = |si /so |, so the transverse magnification of the pinhole camera is, by definition,
MT =
yi
s
=− i ,
yo
so
which is identical with the expression for MT for a lens. Note the negative sign we added here
to reflect the fact that the image is inverted. This formula tells us that the magnification of
the pinhole camera is proportional to si , the pinhole-film plane distance, which is about the
same as the length of the camera box. To obtain a larger image, just extend the length of the
camera body accordingly. (Many commercial pinhole cameras do come with an adjustable body
length.)
24.40
First, use Eqs.(24.8) and (24.9) to find si : MT = yi /yo = −si /so , and so
si = −
yi so
(−1.50 cm)(60.0 cm)
=−
= +30.0 cm .
yo
3.00 cm
(Note that si > 0 since the image is real.) Now find f from Eq. (24.4):
f=
1
1
+
so
si
−1
=
1
1
+
60.0 cm 30.0 cm
−1
= +20.0 cm .
CHAPTER 24
821
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
24.41
Let the magnitude of each radius of the equi-convex lens be R, then R1 = −R2 = R, and the
focal length f of the lens satisfies Eq. (24.2):
1
= (nl − 1)
f
1
1
−
R1
R2
= (nl − 1)
1
1
−
R −R
=
2(nl − 1)
.
R
To find R we need to know f , which can be computed from Eq.(24.4) with the values of so and si .
Since any real image projected by a convex lens is always behind the lens the separation between
the object and the image is d = so + si , which we know to be 0.60 m. Also, the magnification
is MT = yi /yo = −si /so , which gives si = −(yi /yo )so = − (−25 cm/5.0 cm) so = 5.0so . Note
that yi is negative, since any real image projected by a single convex lens is always inverted.
The two last equations above give so = 0.10 m and si = 0.50 m, which we plug into Eq. (24.4),
along with the expression we obtained above for 1/f : 1/f = 2(nl − 1)/R = 1/so + 1/si , and
solve for R:
R=
2(nl − 1)so si
2(1.50 − 1)(0.10 m)(0.50 m)
=
= 0.083 m = 8.3 cm .
so + si
0.10 m + 0.50 m
24.42
First, use the result of Problem (24.15) to find the focal length fw of the lens underwater:
fw =
nl − 1
nl /nw − 1
fa =
1.56 − 1
(10 cm) = 32.383 cm .
1.56/1.33 − 1
Now apply Eq. (24.4), with so = 100 cm and f = fw , for the fish, to find the corresponding
value of si :
s f
(100 cm)(32.283 cm)
si = o w =
= +48 cm ,
so − fw
100 cm − 32.283 cm
i.e., the image of the fish is a distance 48 cm behind the lens.
24.43
The positive lens (with focal length f = 60 cm) projects the object, which is the picture on
the TV screen, onto the wall, where the image of the TV screen is located. So the distance d
between the TV screen and the wall is that between the object and the image: d = so + si .
Now let’s find so and si . Since the picture is enlarged 3 times, the transverse magnification is
[see Eq. (24.7)] MT = −si /so = −3. Here the minus sign corresponds to the fact that, being
real, the image on the wall must inverted. The last equation can be rewritten as si = 3so ,
which we plug into Eq. (24.4), along with f = 60 cm:
1
1
1
1
1
4
1
=
+
=
+
=
,
=
f
0.60 m
so
si
so
3so
3so
822
CHAPTER 24
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
which gives so = 4(0.60 m)/3 = 0.80 m. Thus si = 3so = 3(0.80 m) = 2.4 m. The distance
between the TV screen and the wall is then
d = so + si = 0.80 m + 2.4 m = 3.2 m .
The primary benefit of using a large lens is that more light from the TV screen can be collected
by the lens and projected onto the wall, resulting in a brighter image. Since the image on the
wall is inverted (with respect to the picture on the TV screen), the TV set has to be placed
upside down, if the final image on the wall is to appear normal (i.e., right-side-up).
24.44
Apply Eq. (24.4), with si = so = 100 cm, to obtain
−1 −1
1
1
1
1
f=
+
=
+
= 50.0 cm .
so
si
100 cm 100 cm
Thus by definition the power of the lens is
D=
1
1
= 2.00 m−1 = 2.00 D .
=
f
50.0 cm
Note that the unit for the power is the diopter (D): 1 D ≡ 1 m−1 .
24.45
For the first lens so1 = 20.0 m and f1 = 0.50 m, so Eq. (24.4) gives the location of si1 of the
first image (denoted as I1 in the ray diagram): 1/f1 = 1/so1 + 1/si1 , or
si1 =
so1 f1
(20.0 m)(0.50 m)
=
= 0.512 8 m ,
so1 − f1
20.0 m − 0.50 m
which puts it a distance so2 = 2.00 m − 0.512 8 m = 1.487 m in front of the second lens of focal
length f2 (= 1.00 m). The location of the final image (I2 ) is then
si2 =
so2 f2
(1.487 m)(1.00 m)
=
= 3.1 m ,
so2 − f2
1.487 m − 1.00 m
i.e., it is 3.1 m behind (i.e., to the right of) the second lens. It is real, right-side-up, and minified,
as shown in the ray diagram below.
L1
L2
to object (height y)
y
f1
f1
I1
f2
f2
I2 (final image)
CHAPTER 24
823
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
24.46
With the previous problem in mind, this time so1 = 50.0 cm and f1 = 25.0 cm, hence for the
first image (I1 )
s f
(50.0 cm)(25.0 cm)
si1 = o1 1 =
= 50.0 cm ,
so1 − f1
50.0 cm − 25.0 cm
and so so2 = 130 cm − 50.0 cm = 80.0 cm for the second lens, with f2 = 40.0 cm. Thus
si2 =
so2 f2
(80.0 cm)(40.0 cm)
=
= 80.0 cm ,
so2 − f2
80.0 cm − 40.0 cm
meaning that the final image (I2 ) is located a distance 80.0 cm behind the second lens. The ray
diagram is shown below.
L2
L1
f1
f1
f2
I1
f2
I2 (final image)
24.47
Sunlight comes in parallel, so for the first lens so1 = ∞, and si1 = f1 = 50.0 cm; which means
that, for the second lens (with f2 = 60.0 cm), so2 = 170 cm − 50.0 cm = 120 cm. Thus for the
final image (I2 )
s f
(120 cm)(60.0 cm)
si2 = o2 2 =
= 120 cm ,
so2 − f2
120 cm − 60.0 cm
meaning that the final image of the Sun is at 120 cm behind the second lens. This is where we
should place the screen. The ray diagram is shown below.
L2
L1
parallel light rays
from the Sun
I1
f1
I2 (final image)
f2
f2
824
CHAPTER 24
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
24.48
According to Eq. (24.16) the total power of two lenses in contact is D = D1 + D2 . In this case
D1 = −1.0 D (for the negative lens) and D = −0.75 D, so the power of the liquid lens is
D2 = D − D1 = −0.75 D − (−1.0 D) = +0.25 D .
24.49
The power D is defined as 1/f , which in turn is given by Eq. (24.4). In this case n = 3/2,
R1 = +100 cm = −R2 (for an equiconvex lens); so
1
1
1
1
3
1
D = = (nl − 1)
−
−
=
= +1.00 D .
−1
f
R1
R2
2
1.00 m −1.00 m
24.50
The effective focal length f of two lenses in contact with each other is given by Eq. (24.14):
1/f = 1/f1 + 1/f2 . Plug in f1 = −15.0 cm and f2 = +24.0 cm and solve for f :
f=
1
1
+
f1
f2
−1
=
1
1
+
−15.0 cm 24.0 cm
−1
= −40.0 cm .
The combination is therefore equivalent to a negative lens with a focal length of 40.0 cm.
24.51
Since both lens are positive, adding the second lens will focus the laser beam even faster than
before, resulting in a shorter focal length. So the parallel beam will be focused closer to the
lenses and allowed to spread out more before reaching the screen, causing the size of the blotch
to increase. Since f1 = f2 = 20 cm the focal length f of the two-lens combination is [see
Eq. (24.14)]
−1 −1
1
1
1
1
f=
+
=
+
= 10 cm .
f1
f2
20 cm 20 cm
24.52
Suppose that the three lenses are lined up from left to right, starting from the first lens. Now
consider a parallel beam incident from left through the first lens, which focuses the beam a
distance f1 = 10 cm behind it. Since the second lens is placed a distance 30 cm from the first
one this focused light happens to be a distance 30 cm − 10 cm = 20 cm, or one focal length, in
front of the second lens, whose focal length is f2 = 20 cm. Thus the second lens re-collimates
the light beam, which is subsequently focused again by the third lens, a distance f3 = 5.0 cm
behind it.
CHAPTER 24
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
825
[In general, we need to apply Eq. (24.4) successively three times, once for each lens, to locate
the final image. This particular case is simple since we are dealing with a parallel beam of light,
and the distance between the first two lenses happens to be f1 + f2 .]
24.53
First, combine the first and the second lenses to find the equivalent focal length f12 of the
two-lens combination by applying Eq. (24.14): 1/f12 = 1/f1 + 1/f2 . Now apply Eq. (24.14)
again to combine this with the remaining (third) lens to find the focal length f of the three-lens
system:
1
1
1
1
1
1
1
1
1
+
=
+
+
=
+
+
= 0.25 cm−1 ,
=
f
f12
f3
f1
f2
f3
10 cm −20 cm 5.0 cm
so f = (1/0.25) cm = +4.0 cm. Since f > 0 the lens system is equivalent to a single positive
lens and can form real images.
24.54
The refractive power D is defined as 1/f , which is given by Eq. (24.4):
D=
1
1
1
1
1
+
=
+
= 4.0 m−1 = +4.0 D .
=
f
so
si
1.00 m 0.33 m
24.55
We want the corrective lens to make an object positioned in the normal near-point (25.4 cm)
to appear at 100 cm in front of the lens so it can be viewed clearly. This means that, for
so = 25.4 cm = 0.254 m, we must have si = −100 cm = −1.00 m. Note the negative sign here,
which corresponds to the fact that the image is on the same side as the object. The desired
value of the refractive power of the lens to be prescribed is then
D=
1
1
1
1
1
+
=
+
= 2.94 m−1 = +2.94 D .
=
f
so
si
0.254 m −1.00 m
24.56
This problem is similar to the previous one. In this case the corrective lens should bring the
image of an object at infinity to the far-point, where it can be seen clearly. So for so = +∞
we must have si = −100 cm, or −1.00 m, where the negative sign again corresponds to the fact
that the image is on the same side as the object. The lens to be prescribed should then have a
refractive power of
D=
1
1
1
1
1
+
=
= −1.00 m−1 = −1.00 D .
=
+
f
so
si
∞ −1.00 m
826
CHAPTER 24
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
24.57
Without the corrective lens the person can only clearly see as far as the far-point. What the
lens does is that it brings the image of an object at infinity to the far point, where it can be
viewed clearly. Thus for so = ∞ we must have si = −L, where L is the distance between the
lens and the far-point. Now apply Eq. (24.4), 1/f = 1/so + 1/si = 1/so − 1/L, and solve for
L:
1
1
L=
=
= 5.0 m ,
1/so − 1/f
1/∞ − 1/(−5.0 m)
i.e., the far-point is 5.0 m in front of the lens.
24.58
According to the discussion in the text on the compound microscope the total angular magnification MA is the product of the magnifications of the objective (MTO ) and the eyepiece
(MAE ):
MA = MTO MAE = 20 × 5 = 100× .
24.59
By the definition of the tube length, the total separation Ltotal between the objective lens and
the eyepiece is the sum of of the focal lengths fO and fE of the two lenses plus the tube length L:
Ltotal = fO +fE +L. In this case the total separation is Ltotal = 10.0 cm, fO = 10 mm = 1.0 cm,
and fE = 30 mm = 3.0 cm; so the tube length is
L = Ltotal − fO − fE = 10.0 cm − 1.0 cm − 3.0 cm = 6.0 cm .
24.60
The angular magnification of the telescope is MA = −fO /fE . Let MA = −10× (negative,
since the final image is inverted) and fE = 2.5 cm, and solve for fO :
fO = −MA fE = −(−10)(2.5 cm) = 25 cm .
The length of the scope is then fE + fO = 2.5 cm + 25 cm = 28 cm.
24.61
A telescope consists of an objective lens (O) and an eyepiece (E), with fO > fE . The two
lenses at our disposal have focal lengths 3.0 cm and 1/1.00 m−1 = 1.00 m; so we put the the old
eyeglass lens up front as the objective lens, with fO = 1.00 m, and take the magnifying glass as
the eyepiece, with fE = 3.0 cm. The tube length we’ll need is
L = fO + fE = 1.00 m + 0.030 m = 1.03 m ,
CHAPTER 24
827
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
and the magnification that can be achieved is
MA = −
fO
1.00 m
=−
= −33× .
fE
0.030 m
24.62
(a) Apply the Thin Lens Equation to image 1, the intermediate image: 1/so1 + 1/si1 = 1/f1 ,
to obtain
s f
(15.0 cm)(10.0 cm)
si1 = o1 1 =
= 30.0 cm .
so1 − f1
15.0 cm − 10.0 cm
(b) The magnification is MT1 = −si1 /so1 = −(30.0 cm)/15.0 cm = −2.0.
(c) The image is real (as si1 > 0), inverted (as MT1 < 0), and magnified size (as |MT1 | > 1).
(d) Image 1 is located at (30.0 cm − 25.0 cm = 5.0 cm behind the second lens and is therefore
a virtual object for that lens: so2 = −5.0 cm < 0.
(e) For the final image (image 2, produced by the second lens) 1/so2 + 1/si2 = 1/f2 , and so
si2 =
so2 f2
(−5.0 cm)(−7.5 cm)
=
= +15 cm .
so2 − f2
−5.0 cm − (−7.5 cm)
(f) The final image is real (as si2 > 0), so it can be projected onto a screen.
(g)
I2 (final image)
O
f1
f1
f2
f2
L1
I1
L2
(h) The first lens produces an inverted image (as so1 > f1 ) while the second one keeps the
orientation unchanged from the first image (since it is a concave lens). So the final image is
inverted (relative to the object).
828
CHAPTER 24
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
(i) The magnification of the second lens is MT2 = −si2 /so2 = −(15.0 cm)/(−5.0 cm) = 3.0.
(j) MT = MT1 MT2 = (−2.0)(3.0) = −6.0.
(k) Since |MT | = 6.0 the final image is 6.0 times the size of the object, or 6.0(1.0 cm) = 6.0 cm
tall.
24.63
(a) Apply the Thin Lens Equation to image 1, the intermediate image: 1/so1 + 1/si1 = 1/f1 ,
to obtain
s f
(10.0 cm)(5.0 cm)
si1 = o1 1 =
= 10 cm .
so1 − f1
5.0 cm − 10.0 cm
The magnification is MT1 = −si1 /so1 = −(10.0 cm)/10.0 cm = −1.0. The image is real (as
si1 > 0), inverted (as MT1 < 0), and life size (as |MT1 | = 1.0).
(b) and (c) Image 1 is located at (10.0 cm − 4.0 cm = 6.0 cm behind the second lens and is
a virtual object for that lens: so2 = −6.0 cm. Now 1/so2 + 1/si2 = 1/f2 , and so for the final
image (image 2)
s f
(−6.0 cm)(−3.0 cm)
si2 = o2 2 =
= −6.0 cm .
so2 − f2
−6.0 cm − (−3.0 cm)
The magnification of image 2 (measured relative to image 1, which serves as the object for
image 2) is MT2 = −si2 /so2 = −(−6.0 cm)/(−6.0 cm) = −1.0. The final image is virtual (as
si2 < 0), with a total magnification of MT = MT1 MT2 = (−1.0)(−1.0) = +1.0. It is upright
(as MT > 0) and life size (as |MT | = 1.0).
(d) Since MT = 1.0 the final image has the same size as the object, or 0.30 cm tall.
(e) See Fig. P63 in the textbook.
24.64
First, locate the image projected by the first lens. We have so1 = 30 cm and f1 = +0.30 m =
+30 cm, so from Eq. (24.4)
1
1
1
1
1
=
−
=
−
= 0,
si1
so1
f1
30 cm 30 cm
or si1 = ∞, which means that the light rays from the object becomes parallel after passing
through the first lens: so2 = −∞. The location of the final image can then be obtained from
1/si2 = 1/so2 − 1/f2 , or
si2 =
1
1
=
= −0.20 m ,
1/f2 − 1/so2
1/(−0.20 m) − 1/(−∞)
i.e., the final image is located a distance 0.20 m in front of the second lens, coinciding with
its left focal point (assuming that lens 1 is to the left of lens 2). Since the two lens are 10 cm
apart the final image is also 0.20 m − 10 cm = 10 cm in front of (i.e., to the left of) the first
CHAPTER 24
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
829
lens. It is virtual, with a final transverse magnification MT which equals the product of the
magnifications of the two lenses:
si1
si2
∞ 2
−0.20 m
MT = MT1 MT2 = −
−
= −
=+ ,
−
so1
so2
30 cm
−∞
3
so it is right-side-up and minified to 2/3, or 67%, of the original size of the object.
24.65
The image produced by the first lens is real, so it is 1.0 m behind that lens. Now, the second lens
is 90 cm behind the first one, so the image projected by the first lens is at 1.00 m −90 cm = 10 cm
behind the second lens: so2 = −10 cm. Also, the final image is located 10 cm beyond the first
image, so si2 = 10 cm + 10 cm = 20 cm. Plug the values of both so2 and si2 into Eq. (24.4) to
find f2 , the focal length of the second lens:
f2 =
so2 si2
(−10 cm)(20 cm)
=
= −20 cm .
so2 + si2
−10 cm + 20 cm
Note that f2 < 0, since the second lens is negative. The magnification of the second lens is
MT2 = yi2 /yo2 = −si2 /si1 , so the diameter of the final image of the clock is
yo2 si2 (20 cm)(20 cm)
=
|yi2 | = −
= 40 cm .
so2 10 cm
24.66
For the top portion of the glasses which corrects his near-sightedness, a negative lens of focal
length f1 must be used to cast the image of an object at infinity to the far-point, a distance
3.00 m in front of the cornea, or 3.00 m − 2.0 cm = 2.98 m in front of the glasses. This means
that for so = ∞, si = −2.98 m. Note the negative sign here, since the image is on the same
side of the lens as the object. Plug the values of so and si into Eq. (24.4) to find the desired
power D1 of the (negative) lens:
D1 =
1
1
1
1
1
+
=
+
=
= −0.34 m−1 = −0.34 D .
f1
so
si
∞ −2.98 m
Now consider the lower portion of the glasses which corrects the far-sightedness by bringing
the image of an object placed at a distance dn (= 25.4 cm) in front of the eyes out to the
near-point, which is a distance 45 cm from the eyes, or 45 cm − 2.0 cm = 43 cm from the lens.
So for so = 25.4 cm − 2.0 cm = 23.4 cm we must have si = −43 cm, where the negative sign is
again due to the fact that the image is on the same side of the lens as the object. The desired
power D2 of the (positive) lens is then
D2 =
1
1
1
1
1
=
+
=
+
= +1.9 m−1 = +1.9 D .
f2
so
si
0.234 m −0.43 m
830
CHAPTER 24
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
24.67
(a) Suppose the person wears a pair of glasses with negative lenses of focal length f to correct
his near-sightedness. The lenses help bring the image of an object infinitely far away to the
far-point in front of the eyes so it can be viewed clearly, at 1.00 m − 15.0 mm = 0.985 m from
the lenses. So for so = ∞ we must have si = −0.985 m, where the negative sign is again due
to the fact that the image is on the same side of the lens as the object. The desired power D
of the lenses is then
D=
1
1
1
1
1
=
+
+
=
= −1.02 m−1 = −1.02 D .
f
so
si
∞ −0.985 m
(b) Suppose that the new near-point is a distance dn in front of the eyes, or dn − 15.0 mm in
front of the glasses. An object placed at that point will have its image formed by the glasses a
distance dn = 25.0 cm in front of the eyes — the closest distance from the eyes where an image
can be seen clearly. The distance between this image and the glasses is dn − 15.0 mm. So for
so = dn − 15.0 mm we must have si = −(dn − 15.0 mm) = −(25.0 cm − 15.0 mm) = −23.5 cm.
Plug the values of f , so and si into Eq. (24.4):
D = −1.02 m−1 =
1
1
1
1
1
+
= +
.
=
f
so
si
dn − 15.0 mm −23.5 cm
Solve for dn to obtain dn = 0.324 m = 32.4 cm, which is not terribly inconvenient for reading.
24.68
A microscope consists of an objective lens (O) and an eyepiece (E), with fE > fO > 0. So
fO = 2.0 mm and fE = 2.0 cm. Now locate the position of the image formed by the objective
lens. Plug so = 2.5 mm and fO = 2.0 mm into Eq. (24.4) and solve for si :
si =
fO so
(2.0 mm)(2.5 mm)
=
= 10 mm .
so − fO
2.5 mm − 2.0 mm
The separation of the lenses should then be si + fE = 10 mm + 2.0 cm = 3.0 cm.
Now the magnifications. For the objective
MTO = −
si − fO
10 mm − 2.0 mm
=−
= −4×
fO
2.0 mm
(refer to the discussion in the text following the formula MA = MTO MAE ), and for the eyepiece
MAE =
dn
25.4 cm
=
= 12.7× ;
fE
2.0 cm
so the combined angular magnification of the microscope is
MA = MTO MAE = (−4)(12.7) = −51× .
CHAPTER 24
831
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
24.69
(a) The total magnification is the product of MAE and MTO :
MA = MTO MAE = (−20)(10) = −200× ,
or, to two significant figures, (2.0 × 102 )×.
(b) For the eyepiece MAE = dn /fE , so
fE =
dn
254 mm
=
= 25.4 mm .
MAE
10
For the objective MTO = −L/fO , where L = 160 mm is the standard tube length, so
fO = −
L
160 mm
=−
= 8.0 mm .
MTO
−20
(c) For the image formed by the objective si = L + fO , and so from 1/fO = 1/so + 1/si =
1/so + 1/(L + fO ) we may solve for so :
so =
fO (L + fO )
f (L + fO )
(8.0 mm)(160 mm + 8.0 mm)
= O
= 8.4 mm ,
=
(L + fO ) − fO
L
160 mm
i.e., the object is located 8.4 mm in front of the objective lens.
24.70
Refer to the figure to the right, which shows the four surfaces
of the two lenses. Here R12 = −R21 in accordance with the
problem statement. We are given f , the focal length of the
two-lens system, and f2 , that of L2 . This gives an equation for
f1 , the focal length for L1 , per Eq.(24.14): 1/f = 1/f1 +1/f2 ,
so
L1
L2
R12
R22
R11
R21
1
1
1
1
1
= −
=
−
= 4.0 D .
f1
f
f2
0.50 m −0.50 m
Now, if R11 = R then R12 = −R, as L1 is equi-convex. Eq. (24.2) then gives
1
= (nl1 − 1)
f1
1
1
−
R −R
=
2(nl1 − 1)
.
R
Equate the two expressions above for 1/f1 and plug in nl1 = 1.50 to obtain 4.0 D = 2(1.50 −
1)/R, or R = 0.25 m.
832
CHAPTER 24
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
Now find the remaining unknown radius, R22 , by applying 1/f2 = (nl2 − 1)(1/R21 − 1/R22 ).
Plug in f2 = −0.50 m, nl2 = 1.55, and R21 = R12 = −R = −0.25 m, and solve for R22 :
R22 =
Rf2 (1 − nl2 )
(0.25 m)(−0.50 m)(1 − 1.55)
=
= −2.8 m .
R − f2 (1 − nl2 )
0.25 m − (−0.50 m)(1 − 1.55)
So finally R11 = 0.25 m, R12 = R21 = −0.25 m, and R22 = −2.8 m.
24.71
Before prescribing a new pair of glasses for her we need to know the location of her new nearpoint, which is right at the image formed by the old glasses held at 80 cm from her eyes. When
she holds the book 80 cm away from her eyes it is at 80 cm − 2.0 cm = 78 cm in front of the
glasses. Set D = 1/f = +2.0 D and so = 78 cm = 0.78 m to find 1/si : D = 1/f = 1/so +1/si ,
or
1
1
1
=D−
= 2.0 D −
= 0.718 D .
si
so
0.78 m
The new near-point is a distance si from the eyeglasses. Now, the new glasses should make the
image of any object placed at the new near point appear to be a distance dn = 0.254 m from
the eyes, or 0.254 m − 0.020 m = 0.234 m in front of the eyeglasses. Thus for so = si we must
have si = −0.234 m. The desired power D of the new lens is then
D =
1
1
1
1
+ =
+
= 0.718 D − (−4.27 D) = +5.0 D .
so
si
si
−0.234 m
24.72
To correct the person’s near-sightedness, a negative lens of focal length f must be used to cast
the image of an object at infinity to the far-point, a distance 200 cm in front of the cornea, or
2.00 m − 2.0 cm = 1.98 m in front of the glasses. This means that for so = ∞, si = −1.98 m,
where the negative reflects the fact that the image is on the same side of the lens as the object.
Plug the values of so and si into Eq. (24.4) to find the desired power D of the lens:
D=
1
1
1
1
1
=
+
+
=
= −0.505 D .
f
so
si
∞ −1.98 m
The difference in power between the eyeglasses and the contact lens stems from the fact that
the contact is worn right on the cornea, so the 2.0-cm distance between the cornea and the
glasses is eliminated, and the value of si should just be 200 cm instead of 198 cm for so = ∞.
The power of the contact should then be
1
1
+
= −0.500 D ,
∞ −2.00 m
which corresponds to a focal length of −2.00 m. From the discussion above we see that the far
point in either case is 200 cm in front of the cornea — that’s how the powers are chosen.
Dcontact =
CHAPTER 24
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
833
24.73
According to Eq. (24.18) the focal length f of a spherical mirror is related to its radius of
curvature R via f = −R/2. Plug in R = −66.0 cm to obtain
f =−
R
1
= − (−66.0 cm) = +33.0 cm .
2
2
Note that f > 0, as the mirror is concave.
24.74
The radius of curvature of a convex mirror is positive, so R = +0.50 m. Thus from Eq. (24.18)
the focal length of the mirror is
f =−
R
1
= − (0.50 m) = −0.25 m .
2
2
Note that f < 0, as the mirror is convex.
24.75
The polished steel ball acts as a spherical convex mirror with R = 12 (50 cm) = 25 cm. Thus its
focal length is
R
1
f = − = − (25 cm) = −13 cm .
2
2
24.76
Use a concave mirror of radius |R|, and place the candle a distance |R|/2, or 30 cm, in front of
its vertex. This gives |R| = 2(30 cm) = 60 cm. The location of the image satisfies Eq. (24.19),
1/so + 1/si = −2/R. Note that here R = −|R| = −60 cm < 0 since it is a concave mirror.
Set so = |R|/2 = −R/2 and solve for so :
1
2
2
1
1
=− −
=− −
= 0,
si
R so
R −R/2
or si = ∞, which means that the image will be focused at infinity, so the reflected beam must
be parallel — which is what we wanted. The ray diagram is shown below.
R
f
834
CHAPTER 24
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
24.77
We are given so = 200 cm and si = +400 cm, and we wish to find the value of f for the mirror.
This calls for Eqs. (24.20) and (24.21), 1/so + 1/si = −2/R = 1/f , which we solve for f :
f=
1
1
=
= +133 cm .
1/so + 1/si
1/200 cm + 1/400 cm
Note that f > 0 since the mirror is concave.
24.78
First, locate the image by finding si from Eq. (24.19):
si =
1
1
=
= +60 cm ,
−2/R − 1/so
2/60 cm − 1/60 cm
where the positive sign indicates that the image is at 60 cm in front of the vertex of the mirror,
on the same side as the object itself. It is real, life-sized, and inverted, as shown in the ray
diagram below.
2f
f
24.79
(a) We are given so = 30 cm = 0.30 m for the location of the object and si = 9.0 m for the
location of the image. Note that here si > 0 since the image is projected onto a screen, which
must be on the same side of the mirror as the object — no light rays from the object can reach
behind the mirror. These data lead to R, the radius of curvature of the mirror, via Eq. (24.19):
1/so + 1/si = −2/R, or
R=−
2
2
=−
= −0.58 m .
1/so + 1/si
1/0.30 m + 1/9.0 m
Note that R < 0 since the mirror is concave.
(b) The transverse magnification is given by Eqs. (24.8) and (24.9) as MT = yi /yo = −si /so ,
where yi and yo are the transverse sizes of the image and the object, respectively. To find yi ,
plug yo = 5.0 cm = 0.050 m, along with si = 9.0 m and so = 0.30 m, into the formula for MT
above and solve for yi :
yi = −
yo si
(0.050 m)(9.0 m)
=−
= −1.5 m .
so
0.30 m
CHAPTER 24
835
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
The image is therefore 1.5 m tall and inverted (as indicated by the minus sign in front of yi ).
It is of courses real, since it can be projected onto a screen.
24.80
To cast a real image onto a screen, the mirror must be concave; so R = −0.20 m and f =
−R/2 = −(−0.20 m)/2 = +0.10 m = +10 cm. Also, the screen is 1.10 m from the vertex of
the mirror so si = +1.10 m. Now use Eq. (24.19) to find so :
so = −
1
1
=−
= 0.11 m = 11 cm ,
2/R + 1/si
2/(−0.20 m) + 1/1.10 m
i.e., the candle is to be placed 11 cm away
from the vertex of the mirror. The image
is real (as so > f > 0), with a transverse magnification of MT = −si /so =
−1.10 m/0.11 m = −10, so it is inverted
and enlarged to 10 times the original size
of the candle, as shown in the ray diagram to the right.
f
24.81
From the hint given in the problem statement, we take R = +20 cm and obtain f = −R/2 =
−100 cm/2 = −50 cm for the concave mirror. Plug this value for the focal length, along with
so = 20 cm, into 1/so + 1/si = 1/f to find
si =
so f
(20 cm)(−50 cm)
=
= −14 cm ,
so − f
20 cm − (−50 cm)
where the negative sign indicates that the image appears to be 14 cm behind the vertex of the
mirror. It is virtual (as no light from your nose can bypass the mirror and reach behind its
vertex), with a transverse magnification of
MT = −
si
−14.3 cm
=−
= +0.71 ,
so
20 cm
so it is right-side-up (as MT > 0) and minified to about 71% the original size of your nose, as
shown in the ray diagram in the next page.
836
CHAPTER 24
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
f
24.82
Apply Eq. (24.19), 1/so + 1/si = −2/R, to locate the image of the mouse. Here so = 100 cm
and R = 12 (80.0 cm) = 40.0 cm, so
si =
1
1
=
= −16.7 cm ,
−1/so − 2/R
−1/100 cm − 2/40.0 cm
meaning that the image is located at 16.7 cm behind the vertex of the mirror. It is virtual,
right-side-up, and minified — see the ray diagram below.
f
24.83
With the previous problem in mind, this time so = 0.10 m and R = 1.00 m, so
si =
1
1
= −0.083 m = −8.3 cm ,
=
−1/so − 2/R
−1/0.10 m − 2/1.00 m
CHAPTER 24
837
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
i.e., your image appears to be 8.3 cm behind the glasses, whose outer surface acts like a reflecting
spherical mirror. The image is virtual, right-side-up, and minified, as shown in the ray diagram
below.
f
24.84
The focal length of the convex mirror in question is f = −R/2 = −(−8.0 mm)/2 = −4.0 mm.
Since so = 20 cm, with the previous problem in mind we have
si =
so f
(20 cm)(−0.40 cm)
=
= −0.39 cm = −3.9 mm ,
so − f
20 cm − (−0.40 cm)
where the negative sign indicates that your image appears to be 3.9 mm behind the cornea. It
is virtual, with a transverse magnification of MT = −si /so = −(−0.392 cm)/20 cm = +0.020,
so it is right-side-up (as MT > 0) and minified to about 2% of your original size.
24.85
(a) Since the mirror is concave f > 0. Now, si = 90.0 cm and so =
This gives f , the focal length, via Eq. (24.19): 1/so + 1/si = 1/f , or
f=
1
2 (90.0 cm)
2
1
≈−
= 30.0 cm .
1/si + 1/so
1/(90.0 cm) + 1/(45.0 cm)
(b) The image has to projected onto the screen so it must be real.
(c) Inverted, as is the case for all real images produced by a concave mirror.
(d) Since |si | > |so | (or, since 2f > si > f ) the image is magnified.
= 45.0 cm.
838
CHAPTER 24
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
(e)
O
f
I
(f) MT = −si /so = −90.0 cm/45.0 cm = −2.0.
(g) Since |MT | = 2.0 the image is 2.0 times the size of the object, or 2.0(4.0 mm) = 8.0 mm
tall.
(h) R = −2f = −2(30.0 cm) = −60.0 cm, it is negative since the mirror is concave.
24.86
(a) Since the mirror is convex R > 0 and f = − 12 R = − 12 (50.0 cm) = −25.0 cm.
(b) Virtual, since it is produced by a convex mirror.
(c) Upright, since it is produced by a convex mirror.
(d) Minified, since it is produced by a convex mirror. As the object approaches the mirror the
image size increases. To get a near-life size image the object has to be essentially right in front
of the mirror.
(e)
I
O
f
2f
(f) Apply Eq. (24.19), 1/so + 1/si = −2/R, and solve for si :
si = −
1
1
≈−
= −28.6 cm ,
2/R + 1/so
2/(−50.0 cm) + 1/(200 cm)
i.e., the image is located 28.6 cm behind the mirror.
(g) MT = −si /so = −(−28.6 cm)/200 cm = +0.143.
(h) Since |MT | = 0.143 the image is 0.143 times the size of the object, or 0.143(100 cm) =
14.3 cm tall.
CHAPTER 24
839
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
24.87
First we need to locate the image of the satellite formed by the mirror, by solving for si from
Eq. (24.19), 1/so + 1/si = −2/R. Assuming that the satellite is directly above in the sky, then
so is approximately the altitude of its orbital: so ≈ 500 km = 5.00 × 105 m. Plug this value
for so , along with R = −1.0 m (negative as the mirror is concave), into the equation above and
solve for si :
si = −
1
1
= +0.50 m ,
≈−
2/R + 1/so
2/(−1.0 m) + 1/(5.00 × 105 m)
so the (real) image of the satellite is located 0.50 m in front of the vertex of the mirror. The
size of the image can then be obtained from the transverse magnification MT in Eqs. (24.8) and
(24.9) (which are identical for both thin lenses and spherical mirrors): MT = yo /yi = −si /so .
Here yo = 2.0 m is the size of the satellite, si = 0.50 m, so = 5.00 × 105 m, which we plug into
the equation and solve for |yi |, the size of the image of the satellite in the telescope:
yo si (2.0 m)(0.50 m) = −
= 2.0 × 10−6 m = 2.0 µm .
|yi | = −
5
s
5.00 × 10 m o
(If you left out the absolute value sign you would obtain yi = −2.0 µm, where minus sign only
indicates that the image is inverted.)
24.88
The magnification of the mirror is yi /yo = −si /so . Here yo = 1.0 m, yi = −1.0 cm (note the
negative sign here, since the image is minified and, therefore, real, which means that it must be
inverted), and so = 10 m. Thus
si = −
so yi
(10 m)(−1.0 cm)
=−
= +10 cm = +0.10 m ,
yo
1.0 m
so the detector should be located a distance 10 cm away from the vertex of the mirror. Plug
this value of si into 1/f = 1/so + 1/si , along with so = 10 m, to find f , the focal length of the
mirror:
s s
(10 m)(0.10 m)
f= o i =
= 0.099 m = 9.9 cm .
so + si
10 m + 0.10 m
The ray diagram is shown below.
ray 1
to object, where rays 1 and 2 originate
ray 2
f
image
840
CHAPTER 24
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
24.89
Since a convex mirror can only produce minified images, we should select a concave one for the
task, and put the object (the tooth) within one focal length from the vertex of the mirror to
form a virtual, enlarged image that is right-side-up.
All that remains to be specified is the radius R of the mirror. To find R, Apply Eq. (24.19),
1/so + 1/si = −2/R, with so = 1.5 cm, along with Eq. (24.7) for the magnification: MT =
−si /so = +2 (positive, as the image is right-side-up), or si = −2so = −2(1.5 cm) = −3.0 cm.
Thus
R=−
2
2
=−
= −6.0 cm .
1/so + 1/si
1/1.5 cm − 1/3.0 cm
24.90
First, find the value of si , given R and so , by applying Eq. (24.19): 1/so + 1/si = −2/R, or
si = −
1
Rso
=−
.
2/R + 1/so
2so + R
The magnification is therefore [see Eq. (24.7)]
MT = −
si
−Rso /(2so + R)
R
=−
=
.
so
so
2so + R
24.91
Use the result of the previous problem, with R = −60 cm (negative because the mirror is
concave) and so = 2.4 m, to find the magnification:
MT =
−0.60 m
R
=
= −0.14× .
2so + R
2(2.4 m) + (−0.60 m)
It is real, inverted, and minified, as shown in the ray diagram below.
f
24.92
The magnification MT was computed in Problem (24.90) as a function of so and R: MT =
R/(2so + R), which we may solve for R: R = 2so MT /(1 − MT ). In this case MT = 0.037
and so = 100 mm, so
CHAPTER 24
841
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
R=
2so MT
2(100 mm)(0.037)
=
= +7.7 mm .
1 − MT
1 − 0.037
Here R > 0, since the shape of the cornea surface is convex.
24.93
Similar to Problem (24.90), apply both Eqs. (24.6) and (24.9): 1/so + 1/si = 1/f and MT =
−si /so . This is an equation set for so and si . To obtain so , for example, eliminate si by
substituting si = −MT so into Eq. (24.19) to obtain 1/so + 1/(−MT so ) = 1/f , which we solve
for so :
f (MT − 1)
so =
.
MT
Now substitute so into the equation for MT to obtain MT = −si /so = −si /[f (MT −1)/MT ] =
−si MT /[f (MT − 1)], or
si = −f (MT − 1) .
24.94
Use the result of Problem (24.90): MT = R/(2so + R), with MT = −0.064 and so = 25 cm,
to solve for R, the radius of curvature of the mirror:
R=
2so MT
2(25 cm)(−0.064)
=
= −3.0 cm .
1 − MT
1 − (−0.064)
Here R < 0, as expected, since the mirror is concave.
24.95
A
A2
f
B2
α
β
B
A1
B1
In the ray diagram above, the youngster (represented by AB ) sees her image in the plane mirror
(A1 B1 ) twice as large as her image in the convex mirror (A2 B2 ). This means that α = 2β , where
842
CHAPTER 24
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
α and β are the angles subtended by the images in the plane and convex mirrors, respectively,
as shown. Since the youngster’s size (1.00 m) is considerably less than the distance between
her and her images, both α and β are small, so α ≈ A1 B1 /AA1 and β ≈ A2 B2 /AA2 , where
A1 B1 = AB = 1.0 m (for plane mirror), AA1 is twice the the separation between her and the
plane mirror, or 2 × 5.0 m = 10 m. These relationships can be combined to yield
β≈
α
1.0 m
A2 B2
A B
= ≈ 1 1 ≈
= 0.050 ,
2
2(10 m)
AA2
2AA1
i.e.,
A2 B2
≈ 0.050 .
AA2
(1)
Now consider the image in the convex mirror. We have AA2 ≈ so + |si | = so − si , where
so = 10 m/2 = 5.0 m. Here we noted that si < 0 since the image is behind the vertex of the
mirror. Thus Eq. (24.19) reads
1
1
1
1
1
+
=
+
= .
so
si
5.0 m 5.0 m − AA2
f
(2)
Finally, the magnification of the image in the convex mirror is
MT =
A2 B2
A B
s
5.0 m − AA2
= 2 2 =− i =−
.
1.0 m
so
5.0 m
AB
(3)
There are three variables, namely A2 B2 , AA2 , and f , in Eqs. (1) through (3) above. Solve them
to obtain A2 B2 = 0.33 m, AA2 = 6.7 m, and f = −2.5 m.
24.96
The focal length of the larger mirror (L) is fL = −R2 /2 = −(−2.0 m)/2 = +1.0 m, so the
image of a star (with so = ∞) formed by the larger mirror is focused 1.0 m in front of (i.e., to the
left of) its vertex, or 1.0 m − 34 m = 0.25 m behind the smaller mirror. So for the smaller mirror
(S) the object is virtual, with so = −0.25 m. Since fS = −RS /2 = −0.60 m/2 = −0.30 m, we
may apply 1/so + 1/si = 1/fS to find the location of the final image of the star:
si =
so fS
(−0.25 m)(−0.30 m)
= +1.5 m ,
=
so − fS
−0.25 m − (−0.30 m)
meaning that it is 1.5 m in front of the vertex of the smaller mirror (or 1.5 m to the right of the
mirror, per Fig. P96). This is where the film plane should be located to capture a sharp image
on the film.
Compared with a single concave mirror, the system has an effective focal length of
feff = 1.5 m +
3
m = +2.3 m .
4
CHAPTER 24
GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT
843
This is clear if you imagine that the light rays reflected from the smaller mirror are reversed in
direction (i.e., they now project to the left) — which obviously does not alter the value of feff .
24.97
The socket is placed a distance so = 2f in front of the vertex of a concave mirror. From
1/f = 1/so + 1/si we find si = 2f , so MT = −si /so = −1. The image of the socket produced
by the mirror is therefore real, inverted, and life-sized, at the same location of the socket itself
— this is what you see.
Since 2f = 1.0 m the radius of curvature R of the mirror is
R = −2f = −1.0 m .
24.98
Rewrite the equation y 2 + (x − R)2 = R2 as (x − R)2 = R2 − y 2 , or x = R ± R2 − y 2 .
Since we are interested in comparing the circle
the origin,
with a parabola whose vertex is at
2
2
we only need to consider the case x = R − R − y , as the other case, x = R + R2 − y 2 ,
corresponds to a mirror whose vertex as at x = 2R, rather than x = 0. (You can find the
location of the vertex by setting y = 0.) Thus
2 1/2 y
2
2
x=R− R −y =R 1− 1−
.
R
If (y/R)2 1 then the term in the square root can be approximated by virtue of binomial
expansion. Recall that
(1 + a)n = 1 + na +
n(n − 1) 2 n(n − 1)(n − 2) 3
a +
a + ··· ,
2!
3!
so with a = −(y/R)2 and n = −1/2
2 1/2
2
4
y
1 y
3 y
1−
=1−
+
+ ··· .
R
2 R
8 R
The higher-order terms can be neglected as (y/R)2 1. Keeping to the order (y/R)2 in the
expansion and plugging the result back into the solution for x, we have
2 1/2 2 y
1 y
y2
x=R 1− 1−
,
≈R 1− 1−
=
R
2 R
2R
which we rewrite as y 2 = 2Rx. Comparing this with y 2 = 4f x, we get R = 2f , or
R
f= .
2
Note that the center of the spherical mirror is located at x = R while the vertex is at x = 0.
If we assume that light rays are incident onto the mirror surface from the positive x-axis, then
R > 0 corresponds to a concave mirror (with f > 0), and R < 0 a convex one (with f < 0).
The virtue of the analysis above is that, near its vertex, a spherical mirror of radius R resembles
a parabolic one with a focal length f = R/2, and vice versa.