804 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT 24 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.
•24.1The focal length increases because the rays are not bent as strongly at the water-glass interface.
•24.2It will be positive. It’s a diverging lens.
•24.3The focal length depends on the index of refraction and that depends on the wavelength.
•24.4They provide an air space in front of the cornea so that the eye can work with its normal amountof refraction at that first interface.
•24.5Form 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.
•24.6The rump in the real, the head in the virtual.
•24.7The gas provides a medium with an index greater than one and so lessens the bending of therays. As it is pumped out, the rays are brought to convergence more strongly and the focalpoint moves toward the lens.
•24.8The rays must have been converging to begin with.
•24.9The radius of curvature is ∞ and so is f . That means the object- and image-distances musthave equal magnitudes. Thus, the magnification is +1.
•24.10A point source emits spherical waves which strike the surface and appear to come from a pointsource at the other focus. Hence the reflected waves are spherical too.
•24.11If 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.
•24.12The focal length is shorter and so the dioptric power is greater.
•24.13The target is at one of the two focii of the hyperboloid and rays reflected from it appear tocome 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 theother focus, F2(E), at the film plane.
•24.14The rays will be more converging than necessary so objects at infinity (and far away) willbe blurry because the eye cannot handle converging rays. Nearer objects with more stronglydiverging rays will be seen clearly.
•24.15The object has a diameter d, where d(1200) = 0.0005 m, d = 4.2× 10−7 m, which is the sameas the wavelength of violet light. We cannot hope to see objects that are smaller than the
probe being used; namely, the wavelength of light. The amount of diffraction will then obscurethe image completely. So the details we wish to observe cannot be finer than λ, which puts apractical limit on the magnification. A 12000× microscope will only make larger blurred imagesshowing no more detail.
•24.16The special flat filament of the lamp is at the center of the curvature of the spherical mirror soas to reflect light back toward the filament which is at the focus of the lens. Thus most of thelight 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.
•24.17The filament is located at one focus of the ellipsoid and the reflected light is made to convergeat the other focus, which also corresponds to the front focal point of the lens combination.
•24.18geometrical optics —– An idealized domain of optics, where the diffraction of light is neg-ligible and all the light travel in straight lines in accordance with the Laws of Reflection andRefraction.
lens —– A ground or molded piece of glass, plastic, or other transparent material with oppositesurfaces either or both of which are curved, by means of which light rays are refracted so thatthey 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, tobend 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 asphere.
paraxial —– Rays that are not far from the central axis and enter only at shallow angles areknown to be paraxial.
thin lens —– Lenses for which the radii of curvature of the surfaces are large compared tothe thickness.
object distance —– The distance between the object and the optical instrument (such as alens or a mirror).
image distance —– The distance between the optical instrument (such as a lens or a mirror)and the image it forms.
focal point —– A point at which rays of light or other radiation converge or from which theyappear 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 rayheading 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 opticalaxis) dimension of the image formed by an optical system to the corresponding dimension ofthe object.
longitudinal magnification —– The ratio of any longitudinal (i.e., parallel to the opticalaxis) dimension of the image formed by an optical system to the corresponding dimension ofthe object.
cornea —– The transparent convex anterior portion of the outer fibrous coat of the eyeballthat 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 vitreoushumor 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 andconnected 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 deviceto 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 onewhich serves as the effective object of the next instrument.
dioptic power —– A measure of the focusing power of a lens, defined as the reciprocal ofthe focal length.
farsightedness —– An abnormal condition of the eye in which vision is better for distantobjects 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 theirimages 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.
24.4The lens is convex since it is thicker in the middle. The focal length f is given by Eq. (24.2),with n
l= 1.58, R1 = 1.00 m, and R2 = ∞ (for the flat surface):
f =[(n
l− 1)
(1
R1
− 1R2
)]−1
=[(1.50 − 1)
(1
1.00 m− 1
∞
)]−1
= 2.0 m .
So the sunlight (with so ≈ ∞) will be focused a distance si = f = 2.0 m behind the lens.
24.5Use Eq. (24.2) for the focal length f , with n
l= 1.58, R1 = +1.00 m, and R2 = −1.00 m:
f =[(n
l− 1)
(1
R1
− 1R2
)]−1
=[(1.58 − 1)
(1
1.00 m− 1
−1.00 m
)]−1
= 0.86 m .
24.6For 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 intoEq. (24.2), along with n
l= 1.5, to find f :
f =[(n
l− 1)
(1
R1
− 1R2
)]−1
=[(1.5 − 1)
(1
2.0 m− 1
−5.0 m
)]−1
= 2.9 m .
24.7From the Lensmaker’s Formula [Eq. (24.2)]
f =[(n
l− 1)
(1
R1
− 1R2
)]−1
=[(1.50 − 1)
(1
0.500 m− 1
0.250 m
)]−1
= −1.0 m .
Note that f < 0 since it’s a concave lens.
24.8Similar to the previous problem, this time R1 = 0.250 m and R2 = 0.500 m, so
24.9The focal length f of the lens is given by the Lensmaker’s formula, 1/f = (n
l−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 theformula to solve for n
l:
nl = 1 +1
f(1/R1 − 1/R2)= 1 +
1(1.60 m) (1/2.00 m − 1/∞)
= 2.25 .
24.10It is clear that f = 2.50 m for the lens, since si = 2.50 m when so = ∞. Now set so = 4.00 mand find si from Eq. (24.4), 1/so + 1/si = 1/f :
si =sof
so − f=
(4.00 m)(2.50 m)4.00 m − 2.50 m
= 6.67 m ,
i.e., the screen must be positioned at 6.67 m behind the lens. Since si > 0 the image is real.
24.11From the problem statement we know that si = ∞ when so = 25.0 cm. Plug these intoEq. (24.4) to obtain f = si = 25.0 cm. Now set so = 150 cm and find the corresponding valueof si from Eq. (24.4):
si =sof
so − f=
(150 cm)(25.0 cm)150 cm − 25.0 cm
= 30.0 cm ,
i.e., the image is at 30.0 cm behind the lens. Again, since si > 0 the image is real.
24.12For a plane wave (with parallel light rays) so = ∞, so f = si = 100 cm. Now replace the valueof so with 200 cm and find the new value of si from Eq. (24.4):
si =sof
so − f=
(200 cm)(100 cm)200 cm − 100 cm
= 200 cm .
Again, since si > 0 the image is real.
24.13Use the Lensmaker’s Formula: 1/f = (n
l− 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 ,
l− 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
nl− 1
+1
R2
)−1
=(
1/50.0 cm + 1/∞1.50 − 1
+1∞
)−1
= 25.0 cm .
Thus the radii of the lens are 25.0 cm and ∞.
24.15In deriving Eq. (24.2), 1/fa = (n
l− 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 istands for air and t for the glass. Taking ni = nair = 1 and nt = n
l, this becomes
sin θt = nlsin θi .
In the current case i stands for water (w), so
sin θt =(
nt
ni
)sin θi =
(n
l
nw
)sin θi .
Comparing this expression with that in the previous case, we find that, to obtain the new focallength fw , we must replace n
lwith n
l/nw when the lens is submerged in water in stead of air.
Thus Eq. (24.2) becomes1fw
=(
nl
nw
− 1) (
1R1
− 1R2
).
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.
Thusfa
fw
=n
l/nw − 1n
l− 1
=0.1250.5
=14
,
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.
(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)
(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 =sof
so − f=
(3.00 m)(1.000 m)3.00 m − 1.000 m
= +1.50 m ,
i.e., the image of the person is located 1.50 m behind the lens. It is real (since si > 0), with amagnification of
MT = − si
so
= −1.50 m
3.00 m= −0.500 ,
i.e., it is half-sized and inverted (since MT < 0). Here we used Eq. (24.7) for the transversemagnification.
24.18The object distance (300 cm) is much greater than the focal length (5.00 cm), and much greaterthan 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 =sof
so − f=
(300 cm)(5.00 cm)300 cm − 5.00 cm
= +5.1 cm ,
i.e., the image is located 5.1 cm behind the lens, with a magnification of
MT = − si
so
= − 5.1 cm
300 cm= −0.017 ,
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 sinceit 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)
f =1
1/so + 1/si
=sosi
so + si
=(12.0 cm)(−4.8 cm)12.0 cm − 4.8 cm
= −8.0 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.
24.20Use 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 =sof
so − f=
(30.0 cm)(10.0 cm)30.0 cm − 10.0 cm
= +15.0 cm ,
i.e., the image of the cat is located 15.0 cm behind the lens. It is real (since si > 0), with amagnification of
MT = − si
so
= −15.0 cm
30.0 cm= −0.500 ,
i.e., it is half-sized and inverted (since MT < 0). Here we used Eq. (24.7) for the transversemagnification.
24.21With the previous problem in mind, we use Eq. (24.4), with so = 100.0 cm and f = 60.0 cm,to find si :
si =sof
so − f=
(100.0 cm)(60.0 cm)100.0 cm − 60.0 cm
= +150 cm ,
i.e., the image of the frog is located 150 cm behind the lens. It is real (since si > 0), with amagnification of
MT = − si
so
= − 150 cm
100.0 cm= −1.50 ,
so the image is 50% larger than the frog itself and inverted (since MT < 0).
24.22Similar to the previous two problems, we again apply Eq. (24.4), 1/so + 1/si = 1/f , withso = 100 cm and f = 50.0 cm, to find si :
si =sof
so − f=
(100 cm)(50.0 cm)100 cm − 50.0 cm
= +100 cm ,
i.e., the image of the flower is located 100 cm behind the lens. It is real (since si > 0), with amagnification of
MT = − si
so
= −100 cm
100 cm= −1.00 ,
so the image is life-sized (since |MT | = 1.00) and inverted (since MT < 0).
24.23The location of the image is a distance si from the lens, where si is (see the previous threeproblems)
si =sof
so − f=
(80.0 mm)(120 mm)80.0 mm − 120 mm
= −240 mm ,
where we plugged in so = 80.0 mm and f = 120 mm. Note that here si < 0, so image is virtualand a distance 240 mm in front of the lens (i.e., on the same side of the lens as the object).
24.24The image produced by a concave lens is always virtual, minified, and right-side-up. To locatethe image of the beacon light, plug in so = 50 m and f = −10 m into Eq. (24.4) and solve forsi :
si =sof
so − f=
(50 m)(−10 m)50 m − (−10 m)
= −8.3 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 andright-side-up.
24.25(a) First, find the location of the image by solving for si from Eq. (24.4), which gives
si =sof
so − f=
(0.75 m)(0.25 m)0.75 m − 0.25 m
= 0.38 m ,
so the image is located a distance 0.38 m behind the lens. It is real (since si > 0), with amagnification 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.
24.26The eyeglass is a positive (convex) lens with f = +2.0 m. The location of the image of the bus isat 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 =sif
si − f=
(2.5 m)(2.0 m)2.5 m − 2.0 m
= 10 m ,
i.e., the bus is a distance 10 m away from the eyeglass.
24.27The 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 shouldbe si , so si = 100 mm. It follows that (see the previous problem)
so =sif
si − f=
(100 mm)(60.0 mm)100 mm − 60 mm
= 150 mm ,
so the bug should be a distance 150 mm away from the camera lens.
24.28With the previous problem in mind, we plug f = +52.0 mm, along with so = 5.00 m, intoEq. (24.4) and solve for si , the desired distance between the lens and the film:
Since the camera is originally focused on a distance object (i.e., so ≈ ∞) the original distancebetween the lens and the film is the same as the focal length, at 52.0 mm. So to re-focus whenthe object is 5.00 m from the lens the lens has to be advanced by 52.5 mm−52.0 mm = 0.5 mm.
24.29When the eye is in the relaxed state we may use Eq. (24.9) to find the magnification of the lensto be MA = dn/f , where dn = 0.254 m is the normal viewing distance (to the near-point) fora standard observer. In this case MA = 2.0×, so the focal length of the lens is
f =dn
MA
=0.254 m
2.0= 0.13 m .
24.30Similar to the previous problem, we apply Eq.(24.9) for the magnification MA , with dn = 25 cmand f = 10 cm:
MA =dn
f=
25 cm
10 cm= 2.5× .
24.31The focal length f of the magnifier is related to its magnification MA via MA = dn/f , wheredn = 0.254 m for a standard observer. Since MA = 8×,
f =dn
MA
=0.254 m
8= 0.03 m = 3 cm ,
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 .
i.e., the image of the candle is located 20.0 cm to the left of the lens.
(g)
MT = − si
so
= −−20.0 cm
10.0 cm= +2.00 .
(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 thedistance between the image and the lens — it is negative since it is on the same side as theobject.
(b) Use Eq. (24.4), the Gaussian Lens Equation: 1/f = 1/so + 1/si , to obtain
f =1
1/so + 1/si
=sosi
so + si
=(12.0 cm)(−4.0 cm)12.0 cm − 4.0 cm
= −6.0 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) Upright, as is the case for any image formed by a concave lens.
(g)
MT = − si
so
= −−4.0 cm
12.0 cm= +0.33 .
(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.
24.34First, compute the focal length f of the lens from Eq. (24.2):
f =[(n
l− 1)
(1
R1
− 1R2
)]−1
=[(1.5 − 1)
(1
0.60 m− 1
−0.60 m
)]−1
= 0.60 m .
(a) For a life-sized, real image MT = −si/so = −1.0, so so = si , and Eq. (24.4) now becomes1/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 bereal 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 beplaced 1.2 m behind the lens for a sharp image to be focused on it.
(c)
24.35When an object is placed a distance so in front of a convex lens of focal length f a real image isformed a distance si behind the lens (if so > f). Here so , si and f are related via the GaussianLens Equation [Eq. (24.4)]: 1/f = 1/so + 1/si . In our case the object (grasshopper) is located10 cm to the left of the lens while its image is 30 cm to the right of the lens, so so = 10 cm andsi = 30 cm. Plug these data into Eq. (24.4) to find f , the focal length of the lens:
1f
=1so
+1si
=1
10 cm+
130 cm
=4
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 andsolve for si , the new location of the image:
which means that the image is now 3.8 cm to the left of the lens (i.e., on the same side of thelens 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 tothree 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 magnificationis now MT = −si/so = −3.8 cm/2.5 cm = +1.5 > 0, so the new image is right-side-up and ismagnified to 1.5 times the original size.
(c) The two ray diagrams below depict the situation before and after the grasshopper makesthe jump.
24.36For an image to be projected onto a screen it must be real, so si = +10 m. Also, it is enlarged100 times so |MT | = |si/so | = 100, which gives so = si/100 = 10 m/100 = 0.10 m. Plug thevalues of so and si into Eq. (24.4) to find the desired focal length f of the projector lens:
f =(
1so
+1si
)−1
=(
10.10 m
+1
10 m
)−1
= 0.099 m = 9.9 cm .
24.37The size of Fred-the-chicken’s face is yo = 5.0 cm, while the size of the corresponding image onthe film is yo = 24 mm. Thus the transverse magnification MT provided by the camera lens is
[see Eqs (24.8) and (24.9)] MT = −si/so = yi/yo = 0.24 cm/5.0 cm. Now plug in so = 2.0 minto this equation to find si = 0.96 m. The focal length f of the camera then follows fromEq. (24.4):
f =(
1so
+1si
)−1
=(
12.0 m
+1
0.96 m
)−1
= 0.65 m = 6.5 × 102 mm .
24.38Since the Moon is very far away from us we may take so = 3.84 × 108 m ≈ ∞, which yieldssi ≈ 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:
|yi | =∣∣∣∣−siyo
si
∣∣∣∣ =(50 mm)(3.467 × 106 m)
3.84 × 108 m= 0.45 mm .
(If you left out the absolute value sign you would get yi = −0.45 mm, where the minus signonly means that the image of the Moon is inverted.)
24.39Consider 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
yo
= − si
so
,
which is identical with the expression for MT for a lens. Note the negative sign we added hereto reflect the fact that the image is inverted. This formula tells us that the magnification ofthe pinhole camera is proportional to si , the pinhole-film plane distance, which is about thesame as the length of the camera box. To obtain a larger image, just extend the length of thecamera body accordingly. (Many commercial pinhole cameras do come with an adjustable bodylength.)
24.40First, use Eqs.(24.8) and (24.9) to find si : MT = yi/yo = −si/so , and so
si = −yiso
yo
= − (−1.50 cm)(60.0 cm)3.00 cm
= +30.0 cm .
(Note that si > 0 since the image is real.) Now find f from Eq. (24.4):
24.41Let the magnitude of each radius of the equi-convex lens be R, then R1 = −R2 = R, and thefocal length f of the lens satisfies Eq. (24.2):
1f
= (nl− 1)
(1
R1
− 1R2
)= (n
l− 1)
(1R
− 1−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 betweenthe object and the image is d = so + si , which we know to be 0.60 m. Also, the magnificationis MT = yi/yo = −si/so , which gives si = −(yi/yo)so = − (−25 cm/5.0 cm) so = 5.0so . Notethat 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(n
l− 1)/R = 1/so + 1/si , and
solve for R:
R =2(n
l− 1)sosi
so + si
=2(1.50 − 1)(0.10 m)(0.50 m)
0.10 m + 0.50 m= 0.083 m = 8.3 cm .
24.42First, 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
1.56/1.33 − 1
)(10 cm) = 32.383 cm .
Now apply Eq. (24.4), with so = 100 cm and f = fw , for the fish, to find the correspondingvalue of si :
si =sofw
so − fw
=(100 cm)(32.283 cm)100 cm − 32.283 cm
= +48 cm ,
i.e., the image of the fish is a distance 48 cm behind the lens.
24.43The positive lens (with focal length f = 60 cm) projects the object, which is the picture onthe TV screen, onto the wall, where the image of the TV screen is located. So the distance dbetween 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, beingreal, 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:
which gives so = 4(0.60 m)/3 = 0.80 m. Thus si = 3so = 3(0.80 m) = 2.4 m. The distancebetween 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 collectedby the lens and projected onto the wall, resulting in a brighter image. Since the image on thewall is inverted (with respect to the picture on the TV screen), the TV set has to be placedupside down, if the final image on the wall is to appear normal (i.e., right-side-up).
24.44Apply Eq. (24.4), with si = so = 100 cm, to obtain
f =(
1so
+1si
)−1
=(
1100 cm
+1
100 cm
)−1
= 50.0 cm .
Thus by definition the power of the lens is
D =1f
=1
50.0 cm= 2.00 m−1 = 2.00 D .
Note that the unit for the power is the diopter (D): 1 D ≡ 1 m−1.
24.45For the first lens so1 = 20.0 m and f1 = 0.50 m, so Eq. (24.4) gives the location of si1 of thefirst image (denoted as I1 in the ray diagram): 1/f1 = 1/so1 + 1/si1 , or
si1 =so1f1
so1 − f1
=(20.0 m)(0.50 m)20.0 m − 0.50 m
= 0.512 8 m ,
which puts it a distance so2 = 2.00 m− 0.512 8 m = 1.487 m in front of the second lens of focallength f2 (= 1.00 m). The location of the final image (I2) is then
si2 =so2f2
so2 − f2
=(1.487 m)(1.00 m)1.487 m − 1.00 m
= 3.1 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.
I2 (final image)f1f1 I1
L1L2
f2 f2
f1 f2 f2
L2L1
I1 I2 (final image)parallel light raysfrom the Sun
24.46With the previous problem in mind, this time so1 = 50.0 cm and f1 = 25.0 cm, hence for thefirst image (I1)
si1 =so1f1
so1 − f1
=(50.0 cm)(25.0 cm)50.0 cm − 25.0 cm
= 50.0 cm ,
and so so2 = 130 cm − 50.0 cm = 80.0 cm for the second lens, with f2 = 40.0 cm. Thus
si2 =so2f2
so2 − f2
=(80.0 cm)(40.0 cm)80.0 cm − 40.0 cm
= 80.0 cm ,
meaning that the final image (I2) is located a distance 80.0 cm behind the second lens. The raydiagram is shown below.
24.47Sunlight comes in parallel, so for the first lens so1 = ∞, and si1 = f1 = 50.0 cm; which meansthat, for the second lens (with f2 = 60.0 cm), so2 = 170 cm − 50.0 cm = 120 cm. Thus for thefinal image (I2)
si2 =so2f2
so2 − f2
=(120 cm)(60.0 cm)120 cm − 60.0 cm
= 120 cm ,
meaning that the final image of the Sun is at 120 cm behind the second lens. This is where weshould place the screen. The ray diagram is shown below.
24.48According to Eq. (24.16) the total power of two lenses in contact is D = D1 + D2 . In this caseD1 = −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 .
24.49The 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
D =1f
= (nl− 1)
(1
R1
− 1R2
)=
(32− 1
)(1
1.00 m− 1
−1.00 m
)= +1.00 D .
24.50The 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 =(
1f1
+1f2
)−1
=(
1−15.0 cm
+1
24.0 cm
)−1
= −40.0 cm .
The combination is therefore equivalent to a negative lens with a focal length of 40.0 cm.
24.51Since both lens are positive, adding the second lens will focus the laser beam even faster thanbefore, resulting in a shorter focal length. So the parallel beam will be focused closer to thelenses and allowed to spread out more before reaching the screen, causing the size of the blotchto increase. Since f1 = f2 = 20 cm the focal length f of the two-lens combination is [seeEq. (24.14)]
f =(
1f1
+1f2
)−1
=(
120 cm
+1
20 cm
)−1
= 10 cm .
24.52Suppose that the three lenses are lined up from left to right, starting from the first lens. Nowconsider a parallel beam incident from left through the first lens, which focuses the beam adistance f1 = 10 cm behind it. Since the second lens is placed a distance 30 cm from the firstone this focused light happens to be a distance 30 cm − 10 cm = 20 cm, or one focal length, infront of the second lens, whose focal length is f2 = 20 cm. Thus the second lens re-collimatesthe light beam, which is subsequently focused again by the third lens, a distance f3 = 5.0 cmbehind it.
[In general, we need to apply Eq. (24.4) successively three times, once for each lens, to locatethe 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 .]
24.53First, combine the first and the second lenses to find the equivalent focal length f12 of thetwo-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-lenssystem:
1f
=1
f12
+1f3
=1f1
+1f2
+1f3
=1
10 cm+
1−20 cm
+1
5.0 cm= 0.25 cm−1 ,
so f = (1/0.25) cm = +4.0 cm. Since f > 0 the lens system is equivalent to a single positivelens and can form real images.
24.54The refractive power D is defined as 1/f , which is given by Eq. (24.4):
D =1f
=1so
+1si
=1
1.00 m+
10.33 m
= 4.0 m−1 = +4.0 D .
24.55We 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, forso = 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 desiredvalue of the refractive power of the lens to be prescribed is then
D =1f
=1so
+1si
=1
0.254 m+
1−1.00 m
= 2.94 m−1 = +2.94 D .
24.56This problem is similar to the previous one. In this case the corrective lens should bring theimage 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 factthat the image is on the same side as the object. The lens to be prescribed should then have arefractive power of
24.57Without the corrective lens the person can only clearly see as far as the far-point. What thelens does is that it brings the image of an object at infinity to the far point, where it can beviewed clearly. Thus for so = ∞ we must have si = −L, where L is the distance between thelens and the far-point. Now apply Eq. (24.4), 1/f = 1/so + 1/si = 1/so − 1/L, and solve forL:
L =1
1/so − 1/f=
11/∞− 1/(−5.0 m)
= 5.0 m ,
i.e., the far-point is 5.0 m in front of the lens.
24.58According to the discussion in the text on the compound microscope the total angular mag-nification MA is the product of the magnifications of the objective (MTO) and the eyepiece(MAE):
MA = MTOMAE = 20 × 5 = 100× .
24.59By the definition of the tube length, the total separation Ltotal between the objective lens andthe 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 .
24.60The 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 = −MAfE = −(−10)(2.5 cm) = 25 cm .
The length of the scope is then fE + fO = 2.5 cm + 25 cm = 28 cm.
24.61A telescope consists of an objective lens (O) and an eyepiece (E), with fO > fE . The twolenses at our disposal have focal lengths 3.0 cm and 1/1.00 m−1 = 1.00 m; so we put the the oldeyeglass lens up front as the objective lens, with fO = 1.00 m, and take the magnifying glass asthe eyepiece, with fE = 3.0 cm. The tube length we’ll need is
(a) Apply the Thin Lens Equation to image 1, the intermediate image: 1/so1 + 1/si1 = 1/f1 ,to obtain
si1 =so1f1
so1 − f1
=(15.0 cm)(10.0 cm)15.0 cm − 10.0 cm
= 30.0 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 thereforea 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 =so2f2
so2 − f2
=(−5.0 cm)(−7.5 cm)−5.0 cm − (−7.5 cm)
= +15 cm .
(f) The final image is real (as si2 > 0), so it can be projected onto a screen.
(g)
(h) The first lens produces an inverted image (as so1 > f1) while the second one keeps theorientation unchanged from the first image (since it is a concave lens). So the final image isinverted (relative to the object).
(i) The magnification of the second lens is MT2 = −si2/so2 = −(15.0 cm)/(−5.0 cm) = 3.0.(j) MT = MT1MT2 = (−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 cmtall.
24.63
(a) Apply the Thin Lens Equation to image 1, the intermediate image: 1/so1 + 1/si1 = 1/f1 ,to obtain
si1 =so1f1
so1 − f1
=(10.0 cm)(5.0 cm)5.0 cm − 10.0 cm
= 10 cm .
The magnification is MT1 = −si1/so1 = −(10.0 cm)/10.0 cm = −1.0. The image is real (assi1 > 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 isa virtual object for that lens: so2 = −6.0 cm. Now 1/so2 + 1/si2 = 1/f2 , and so for the finalimage (image 2)
si2 =so2f2
so2 − f2
=(−6.0 cm)(−3.0 cm)−6.0 cm − (−3.0 cm)
= −6.0 cm .
The magnification of image 2 (measured relative to image 1, which serves as the object forimage 2) is MT2 = −si2/so2 = −(−6.0 cm)/(−6.0 cm) = −1.0. The final image is virtual (assi2 < 0), with a total magnification of MT = MT1MT2 = (−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.
24.64First, 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)
1si1
=1
so1
− 1f1
=1
30 cm− 1
30 cm= 0 ,
or si1 = ∞, which means that the light rays from the object becomes parallel after passingthrough the first lens: so2 = −∞. The location of the final image can then be obtained from1/si2 = 1/so2 − 1/f2 , or
si2 =1
1/f2 − 1/so2
=1
1/(−0.20 m) − 1/(−∞)= −0.20 m ,
i.e., the final image is located a distance 0.20 m in front of the second lens, coinciding withits left focal point (assuming that lens 1 is to the left of lens 2). Since the two lens are 10 cmapart the final image is also 0.20 m − 10 cm = 10 cm in front of (i.e., to the left of) the first
lens. It is virtual, with a final transverse magnification MT which equals the product of themagnifications of the two lenses:
MT = MT1MT2 =(− si1
so1
)(− si2
so2
)=
(− ∞
30 cm
)(−−0.20 m
−∞
)= +
23
,
so it is right-side-up and minified to 2/3, or 67%, of the original size of the object.
24.65The image produced by the first lens is real, so it is 1.0 m behind that lens. Now, the second lensis 90 cm behind the first one, so the image projected by the first lens is at 1.00 m−90 cm = 10 cmbehind the second lens: so2 = −10 cm. Also, the final image is located 10 cm beyond the firstimage, so si2 = 10 cm + 10 cm = 20 cm. Plug the values of both so2 and si2 into Eq. (24.4) tofind f2 , the focal length of the second lens:
f2 =so2si2
so2 + si2
=(−10 cm)(20 cm)−10 cm + 20 cm
= −20 cm .
Note that f2 < 0, since the second lens is negative. The magnification of the second lens isMT2 = yi2/yo2 = −si2/si1 , so the diameter of the final image of the clock is
|yi2 | =∣∣∣∣−yo2si2
so2
∣∣∣∣ =(20 cm)(20 cm)
10 cm= 40 cm .
24.66For the top portion of the glasses which corrects his near-sightedness, a negative lens of focallength f1 must be used to cast the image of an object at infinity to the far-point, a distance3.00 m in front of the cornea, or 3.00 m − 2.0 cm = 2.98 m in front of the glasses. This meansthat for so = ∞, si = −2.98 m. Note the negative sign here, since the image is on the sameside of the lens as the object. Plug the values of so and si into Eq. (24.4) to find the desiredpower D1 of the (negative) lens:
D1 =1f1
=1so
+1si
=1∞ +
1−2.98 m
= −0.34 m−1 = −0.34 D .
Now consider the lower portion of the glasses which corrects the far-sightedness by bringingthe image of an object placed at a distance dn (= 25.4 cm) in front of the eyes out to thenear-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 isagain due to the fact that the image is on the same side of the lens as the object. The desiredpower D2 of the (positive) lens is then
(a) Suppose the person wears a pair of glasses with negative lenses of focal length f to correcthis near-sightedness. The lenses help bring the image of an object infinitely far away to thefar-point in front of the eyes so it can be viewed clearly, at 1.00 m − 15.0 mm = 0.985 m fromthe lenses. So for so = ∞ we must have si = −0.985 m, where the negative sign is again dueto the fact that the image is on the same side of the lens as the object. The desired power Dof the lenses is then
D =1f
=1so
+1si
=1∞ +
1−0.985 m
= −1.02 m−1 = −1.02 D .
(b) Suppose that the new near-point is a distance d′n
in front of the eyes, or d′n− 15.0 mm in
front of the glasses. An object placed at that point will have its image formed by the glasses adistance dn = 25.0 cm in front of the eyes — the closest distance from the eyes where an imagecan be seen clearly. The distance between this image and the glasses is dn − 15.0 mm. So forso = d′
n−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 =1f
=1so
+1si
=1
d′n− 15.0 mm
+1
−23.5 cm.
Solve for d′n
to obtain d′n
= 0.324 m = 32.4 cm, which is not terribly inconvenient for reading.
24.68A microscope consists of an objective lens (O) and an eyepiece (E), with fE > fO > 0. SofO = 2.0 mm and fE = 2.0 cm. Now locate the position of the image formed by the objectivelens. Plug so = 2.5 mm and fO = 2.0 mm into Eq. (24.4) and solve for si :
si =fOso
so − fO
=(2.0 mm)(2.5 mm)2.5 mm − 2.0 mm
= 10 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
fO
= −10 mm − 2.0 mm
2.0 mm= −4×
(refer to the discussion in the text following the formula MA = MTOMAE), and for the eyepiece
MAE =dn
fE
=25.4 cm
2.0 cm= 12.7× ;
so the combined angular magnification of the microscope is
(a) The total magnification is the product of MAE and MTO :
MA = MTOMAE = (−20)(10) = −200× ,
or, to two significant figures, (2.0 × 102)×.
(b) For the eyepiece MAE = dn/fE , so
fE =dn
MAE
=254 mm
10= 25.4 mm .
For the objective MTO = −L/fO , where L = 160 mm is the standard tube length, so
fO = − L
MTO
= −160 mm
−20= 8.0 mm .
(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)
(L + fO) − fO
=fO(L + fO)
L=
(8.0 mm)(160 mm + 8.0 mm)160 mm
= 8.4 mm ,
i.e., the object is located 8.4 mm in front of the objective lens.
24.70Refer to the figure to the right, which shows the four surfacesof the two lenses. Here R12 = −R21 in accordance with theproblem statement. We are given f , the focal length of thetwo-lens system, and f2 , that of L2 . This gives an equation forf1 , the focal length for L1 , per Eq.(24.14): 1/f = 1/f1 +1/f2 ,so
1f1
=1f− 1
f2
=1
0.50 m− 1
−0.50 m= 4.0 D .
Now, if R11 = R then R12 = −R, as L1 is equi-convex. Eq. (24.2) then gives
1f1
= (nl1 − 1)
(1R
− 1−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 −
So finally R11 = 0.25 m, R12 = R21 = −0.25 m, and R22 = −2.8 m.
24.71Before prescribing a new pair of glasses for her we need to know the location of her new near-point, which is right at the image formed by the old glasses held at 80 cm from her eyes. Whenshe holds the book 80 cm away from her eyes it is at 80 cm − 2.0 cm = 78 cm in front of theglasses. 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
1si
= D − 1so
= 2.0 D − 10.78 m
= 0.718 D .
The new near-point is a distance si from the eyeglasses. Now, the new glasses should make theimage of any object placed at the new near point appear to be a distance dn = 0.254 m fromthe eyes, or 0.254 m − 0.020 m = 0.234 m in front of the eyeglasses. Thus for s′
o= si we must
have s′i= −0.234 m. The desired power D′ of the new lens is then
D′ =1s′o
+1s′i
=1si
+1
−0.234 m= 0.718 D − (−4.27 D) = +5.0 D .
24.72To correct the person’s near-sightedness, a negative lens of focal length f must be used to castthe image of an object at infinity to the far-point, a distance 200 cm in front of the cornea, or2.00 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 =1f
=1so
+1si
=1∞ +
1−1.98 m
= −0.505 D .
The difference in power between the eyeglasses and the contact lens stems from the fact thatthe contact is worn right on the cornea, so the 2.0-cm distance between the cornea and theglasses 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
Dcontact =1∞ +
1−2.00 m
= −0.500 D ,
which corresponds to a focal length of −2.00 m. From the discussion above we see that the farpoint in either case is 200 cm in front of the cornea — that’s how the powers are chosen.
24.73According to Eq. (24.18) the focal length f of a spherical mirror is related to its radius ofcurvature R via f = −R/2. Plug in R = −66.0 cm to obtain
f = −R
2= −1
2(−66.0 cm) = +33.0 cm .
Note that f > 0, as the mirror is concave.
24.74The 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
2= −1
2(0.50 m) = −0.25 m .
Note that f < 0, as the mirror is convex.
24.75The polished steel ball acts as a spherical convex mirror with R = 1
2 (50 cm) = 25 cm. Thus itsfocal length is
f = −R
2= −1
2(25 cm) = −13 cm .
24.76Use a concave mirror of radius |R|, and place the candle a distance |R|/2, or 30 cm, in front ofits 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 :
1si
= − 2R
− 1so
= − 2R
− 1−R/2
= 0 ,
or si = ∞, which means that the image will be focused at infinity, so the reflected beam mustbe parallel — which is what we wanted. The ray diagram is shown below.
24.77We 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/so + 1/si
=1
1/200 cm + 1/400 cm= +133 cm .
Note that f > 0 since the mirror is concave.
24.78First, locate the image by finding si from Eq. (24.19):
si =1
−2/R − 1/so
=1
2/60 cm − 1/60 cm= +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 raydiagram below.
24.79
(a) We are given so = 30 cm = 0.30 m for the location of the object and si = 9.0 m for thelocation of the image. Note that here si > 0 since the image is projected onto a screen, whichmust be on the same side of the mirror as the object — no light rays from the object can reachbehind 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 = − 21/so + 1/si
= − 21/0.30 m + 1/9.0 m
= −0.58 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
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.
24.80To 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 ofthe mirror so si = +1.10 m. Now use Eq. (24.19) to find so :
so = − 12/R + 1/si
= − 12/(−0.20 m) + 1/1.10 m
= 0.11 m = 11 cm ,
i.e., the candle is to be placed 11 cm awayfrom the vertex of the mirror. The imageis real (as so > f > 0), with a trans-verse magnification of MT = −si/so =−1.10 m/0.11 m = −10, so it is invertedand enlarged to 10 times the original sizeof the candle, as shown in the ray dia-gram to the right.
24.81From 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 withso = 20 cm, into 1/so + 1/si = 1/f to find
si =sof
so − f=
(20 cm)(−50 cm)20 cm − (−50 cm)
= −14 cm ,
where the negative sign indicates that the image appears to be 14 cm behind the vertex of themirror. It is virtual (as no light from your nose can bypass the mirror and reach behind itsvertex), with a transverse magnification of
MT = − si
so
= −−14.3 cm
20 cm= +0.71 ,
so it is right-side-up (as MT > 0) and minified to about 71% the original size of your nose, asshown in the ray diagram in the next page.
i.e., your image appears to be 8.3 cm behind the glasses, whose outer surface acts like a reflectingspherical mirror. The image is virtual, right-side-up, and minified, as shown in the ray diagrambelow.
24.84The 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 =sof
so − f=
(20 cm)(−0.40 cm)20 cm − (−0.40 cm)
= −0.39 cm = −3.9 mm ,
where the negative sign indicates that your image appears to be 3.9 mm behind the cornea. Itis 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 = 12 (90.0 cm) = 45.0 cm.
This gives f , the focal length, via Eq. (24.19): 1/so + 1/si = 1/f , or
f =2
1/si + 1/so
≈ − 11/(90.0 cm) + 1/(45.0 cm)
= 30.0 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.
(g) Since |MT | = 2.0 the image is 2.0 times the size of the object, or 2.0(4.0 mm) = 8.0 mmtall.
(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 = − 12R = − 1
2 (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 theimage size increases. To get a near-life size image the object has to be essentially right in frontof the mirror.
(e)
(f) Apply Eq. (24.19), 1/so + 1/si = −2/R, and solve for si :
si = − 12/R + 1/so
≈ − 12/(−50.0 cm) + 1/(200 cm)
= −28.6 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.
24.87First we need to locate the image of the satellite formed by the mirror, by solving for si fromEq. (24.19), 1/so +1/si = −2/R. Assuming that the satellite is directly above in the sky, thenso is approximately the altitude of its orbital: so ≈ 500 km = 5.00 × 105 m. Plug this valuefor so , along with R = −1.0 m (negative as the mirror is concave), into the equation above andsolve for si :
si = − 12/R + 1/so
≈ − 12/(−1.0 m) + 1/(5.00 × 105 m)
= +0.50 m ,
so the (real) image of the satellite is located 0.50 m in front of the vertex of the mirror. Thesize 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 intothe equation and solve for |yi |, the size of the image of the satellite in the telescope:
|yi | =∣∣∣∣−yosi
so
∣∣∣∣ =∣∣∣∣− (2.0 m)(0.50 m)
5.00 × 105 m
∣∣∣∣ = 2.0 × 10−6 m = 2.0 µm .
(If you left out the absolute value sign you would obtain yi = −2.0 µm, where minus sign onlyindicates that the image is inverted.)
24.88The magnification of the mirror is yi/yo = −si/so . Here yo = 1.0 m, yi = −1.0 cm (note thenegative sign here, since the image is minified and, therefore, real, which means that it must beinverted), and so = 10 m. Thus
si = −soyi
yo
= − (10 m)(−1.0 cm)1.0 m
= +10 cm = +0.10 m ,
so the detector should be located a distance 10 cm away from the vertex of the mirror. Plugthis value of si into 1/f = 1/so +1/si , along with so = 10 m, to find f , the focal length of themirror:
24.89Since a convex mirror can only produce minified images, we should select a concave one for thetask, and put the object (the tooth) within one focal length from the vertex of the mirror toform 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 = − 21/so + 1/si
= − 21/1.5 cm − 1/3.0 cm
= −6.0 cm .
24.90First, find the value of si , given R and so , by applying Eq. (24.19): 1/so + 1/si = −2/R, or
si = − 12/R + 1/so
= − Rso
2so + R.
The magnification is therefore [see Eq. (24.7)]
MT = − si
so
= −−Rso/(2so + R)so
=R
2so + R.
24.91Use the result of the previous problem, with R = −60 cm (negative because the mirror isconcave) and so = 2.4 m, to find the magnification:
MT =R
2so + R=
−0.60 m
2(2.4 m) + (−0.60 m)= −0.14× .
It is real, inverted, and minified, as shown in the ray diagram below.
24.92The 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 = 2soMT/(1 − MT). In this case MT = 0.037and so = 100 mm, so
Here R > 0, since the shape of the cornea surface is convex.
24.93Similar 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 bysubstituting si = −MTso into Eq. (24.19) to obtain 1/so +1/(−MTso) = 1/f , which we solvefor so :
so =f(MT − 1)
MT
.
Now substitute so into the equation for MT to obtain MT = −si/so = −si/[f(MT −1)/MT ] =−siMT/[f(MT − 1)], or
si = −f(MT − 1) .
24.94Use 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 =2soMT
1 − MT
=2(25 cm)(−0.064)
1 − (−0.064)= −3.0 cm .
Here R < 0, as expected, since the mirror is concave.
24.95
In the ray diagram above, the youngster (represented by AB) sees her image in the plane mirror
(A1B1) twice as large as her image in the convex mirror (A2B2). This means that α = 2β, where
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 ≈ A1B1/AA1 and β ≈ A2B2/AA2 , where
A1B1 = AB = 1.0 m (for plane mirror), AA1 is twice the the separation between her and theplane mirror, or 2 × 5.0 m = 10 m. These relationships can be combined to yield
β ≈ A2B2
AA2
=α
2≈ A1B1
2AA1
≈ 1.0 m
2(10 m)= 0.050 ,
i.e.,A2B2
AA2
≈ 0.050 . (1)
Now consider the image in the convex mirror. We have AA2 ≈ so + |si | = so − si , whereso = 10 m/2 = 5.0 m. Here we noted that si < 0 since the image is behind the vertex of themirror. Thus Eq. (24.19) reads
1so
+1si
=1
5.0 m+
15.0 m − AA2
=1f
. (2)
Finally, the magnification of the image in the convex mirror is
MT =A2B2
AB=
A2B2
1.0 m= − si
so
= −5.0 m − AA2
5.0 m. (3)
There are three variables, namely A2B2 , AA2 , and f , in Eqs.(1) through (3) above. Solve them
to obtain A2B2 = 0.33 m, AA2 = 6.7 m, and f = −2.5 m.
24.96The focal length of the larger mirror (L) is fL = −R2/2 = −(−2.0 m)/2 = +1.0 m, so theimage of a star (with so = ∞) formed by the larger mirror is focused 1.0 m in front of (i.e., to theleft of) its vertex, or 1.0 m− 3
4 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, wemay apply 1/so + 1/si = 1/fS to find the location of the final image of the star:
si =sofS
so − fS
=(−0.25 m)(−0.30 m)−0.25 m − (−0.30 m)
= +1.5 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 themirror, per Fig. P96). This is where the film plane should be located to capture a sharp imageon the film.
Compared with a single concave mirror, the system has an effective focal length of
This is clear if you imagine that the light rays reflected from the smaller mirror are reversed indirection (i.e., they now project to the left) — which obviously does not alter the value of feff .
24.97The socket is placed a distance so = 2f in front of the vertex of a concave mirror. From1/f = 1/so + 1/si we find si = 2f , so MT = −si/so = −1. The image of the socket producedby 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 .
24.98Rewrite the equation y2 + (x − R)2 = R2 as (x − R)2 = R2 − y2, or x = R ±
√R2 − y2.
Since we are interested in comparing the circle with a parabola whose vertex is at the origin,
we only need to consider the case x = R −√
R2 − y2, as the other case, x = R +√
R2 − y2,corresponds to a mirror whose vertex as at x = 2R, rather than x = 0. (You can find thelocation of the vertex by setting y = 0.) Thus
x = R −√
R2 − y2 = R
{1 −
[1 −
(y
R
)2]1/2}.
If (y/R)2 � 1 then the term in the square root can be approximated by virtue of binomialexpansion. Recall that
(1 + a)n = 1 + na +n(n − 1)
2!a2 +
n(n − 1)(n − 2)3!
a3 + · · · ,
so with a = −(y/R)2 and n = −1/2
[1 −
(y
R
)2]1/2
= 1 − 12
(y
R
)2
+38
(y
R
)4
+ · · · .
The higher-order terms can be neglected as (y/R)2 � 1. Keeping to the order (y/R)2 in theexpansion and plugging the result back into the solution for x, we have
x = R
{1 −
[1 −
(y
R
)2]1/2}≈ R
{1 −
[1 − 1
2
(y
R
)2]}=
y2
2R,
which we rewrite as y2 = 2Rx. Comparing this with y2 = 4fx, we get R = 2f , or
f =R
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, thenR > 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 resemblesa parabolic one with a focal length f = R/2, and vice versa.
