Download Optical Lenses and Devices

Optical Lenses and Devices
21
In the last chapter we learned the fundamental laws that tell us how light behaves when
traveling through a transparent material. Here we apply those to the most important
case of optical lenses, both human-made, for use in optical devices, and naturally
occurring, as in the eye. After learning the basics of imaging using a single simple
lens, we show how the human eye exquisitely functions to allow us to see in color at
extremely high resolution. Two human-made optical devices, the magnifying glass
and the compound microscope, are then examined. These can be fairly well understood with only the tools of geometric optics that we have learned. Understanding the
huge arsenal of new optical microscopies studied in Chapter 23 requires knowledge of
wave optics, presented in the next chapter. There we introduce the wave nature of light
and some of its major consequences.
1. OPTICAL LENSES
In the previous chapter, after introducing reflection and refraction we focused on the
phenomena of imaging from reflections in spherical mirrors and total internal reflection. Here we turn to the portion of the light incident on a transparent glass surface
that is transmitted. Lenses are perhaps the most important of optical devices. For a
glass lens, we know that only about 4% of the incident light at near normal incidence
(in the paraxial approximation, with light traveling nearly along the optic axis) will
be reflected at each boundary with air and so most of the light will be transmitted
after being refracted by the curved surfaces. Special optical antireflection coatings
can even increase the transmitted light closer to 100%. Armed with the law of refraction, we can repeat an analysis similar to that of the last chapter for reflected light to
trace the refracted rays of light and to describe the characteristics of the images
formed by a lens.
Lenses are usually made either of glass or clear plastic and are ground and polished to have spherical surfaces. Several varieties of two basic forms of lenses exist:
converging lenses, those thicker at the center than at the edges, and diverging lenses,
those thinner at the center than at the edges (Figure 21.1). Occasionally lenses are
made with cylindrical or even other surface contours for special purposes; these are
not discussed here. Many common lenses, especially those in cameras, are not simple single pieces of glass, but are compound lenses made by cementing many individual lenses together with a transparent glue that has a similar index of refraction to
that of the glass. We discuss these later, focusing first on simple lenses, those that
have negligible thickness compared to their diameter. These are known as thin lenses
and the equations we introduce are limited to such lenses.
Let’s first consider a double convex lens shown in Figure 21.2. Incident rays parallel to the optic axis will be bent by refraction at both surfaces of the lens toward the
optic axis (Figure 21.2a). For a thin lens there is a common point, the focal point F,
J. Newman, Physics of the Life Sciences, DOI: 10.1007/978-0-387-77259-2_21,
© Springer Science+Business Media, LLC 2008
OPTICAL LENSES
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double convex
at which all these rays cross the optic axis, the straight line passing through the lens
center and traveling perpendicular to both surfaces. The distance f from this point to
the center of the lens is called the focal length of the lens and is the same distance to
either side of the lens. That is, if the lens is rotated 180o about a vertical axis it will
focus light at the same point. The focal length of a thin lens can be shown to be
related to its radii of curvature R1 and R2 on each of its sides and its index of refraction n by the lens-maker’s equation
plano-convex
1
1
1
(n 1)a b .
f
R1 R2
double concave
plano-concave
FIGURE 21.1 Converging (upper)
and diverging (lower) lenses.
(21.1)
Note that this equation defines a single focal length for a lens, regardless of the
side facing the incident light, even if the two radii of curvature are different. In this
equation the radii are taken as positive if the surface is convex and negative if concave (discussed below). Note that a plane surface has an infinite radius of curvature.
Example 21.1 A plano-convex lens of refractive index 1.52 has a radius of curvature of 5 cm. First find its focal length. Suppose that a second plano-convex
lens with the same index of refraction is cemented to the first along their planar
faces. What radius of curvature is needed on this second lens to produce a net
focal length 1/4 the value of the focal length of the first lens?
Solution: According to Equation (21.1) the focal length of the first lens is
[0.52/5]1 9.6 cm. The second lens must have a radius of curvature R, such that
(9.6/4) [0.52(1/5 1/R)]1. Solving for R, we find that R 1.7 cm.
FIGURE 21.2 (a) Refraction at
convex lens surfaces tends to
bend light toward the optic axis.
(b) The focal point of a double
convex thin lens.
For a double convex (or any converging) lens, if an object O is to be imaged in
the lens, we can trace three characteristic rays to locate the image. For the object
shown in Figure 21.3, those three rays from the object arrowhead are: (1) a ray parallel to the optic axis that will be focused through the focal point on the far side of
the lens (in red); (2) a ray passing through the focal point on the same side of the lens
that, on passing through the lens, will be refracted to lie parallel to the optic axis
(in blue); and (3) a ray passing through the center of the lens that will continue undeviated (this occurs because both sides of the lens are essentially parallel; the negligible thickness of the lens eliminates any parallel displacement of the ray; shown in
green). The image of the arrow tip representing the object can be determined by the
common point at which these three rays cross (of course, any two of these will cross
at the image point). Then the entire image I is known since the ray along the optic
axis passes straight through and images the tail of the arrow. Note that in this case the
image is upside down, or inverted, and smaller in size.
In place of raytracing we can derive an equation that will allow us to find the
image location as well as its lateral magnification. With the various distances defined
in Figure 21.4, the derivation assumes paraxial optics (with incoming rays making
small angles with the optic axis). There are two sets of similar right triangles. The first
set (shown in green hatched lines) consists of one formed with the object height and
distance as legs (with hypotenuse OC) and the other with the
image height and distance as legs (and hypotenuse IC). From
the similarity of these we have
F
a
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b
f
s
h
,
h¿
s¿
(21.2)
and a second set of similar triangles (one shown in red hatched
lines and the other having F as a common vertex) yields
OPTICAL LENSES
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f
h
.
h¿
s¿ f
(21.3)
2
f
(21.4)
We can also invert Equation (21.2) to find an expression for the lateral magnification m
m
I
3
Combining the two equations, we have that s/s f/(s f ), and after
cross-multiplying and dividing both sides by the product (s s f ) and simplifying the result (try it!), we find the lens equation
1
1
1
.
s s¿
f
1
O
s¿
h¿
,
s
h
f
FIGURE 21.3 Raytracing with a thin
converging lens. The image lies at
the intersection of the three numbered rays (see text for explanation
of the construction).
(21.5)
where the minus sign was introduced so that an inverted image has a negative magnification, and an erect image has a positive magnification.
In the case worked out above using ray diagrams, the object distance is greater
than the focal length and we can see from Equation (21.4) that then because 1/s 1/f, the image distance s is positive, indicating that the image is real and that the
image will be inverted (because m 0 in that case). If s s the image will be magnified, whereas if s s the image will be smaller than the object. It is easy to see
that the dividing line occurs when s 2f, because then s 2f as well and the image
will be “life-size,” but still inverted. If s 2f, then f s 2f and the image will be
smaller, whereas if f s 2f, the image will be magnified.
The reciprocal of the focal length for a lens is called its power P, where
1
P .
f
(21.6)
Lens power is measured in reciprocal meters which are called diopters
(D). Thus, the shorter the focal length of a lens is, the stronger its power. The
diopter unit is mainly used in coding eyeglass lenses, a topic we return to in the
next section.
At this point, if we generalize Equations (21.4) and (21.5) using a set of sign rules,
these equations are then valid for all thin lenses no matter what the configuration.
These rules are given in Table 21.1. We illustrate their use with a few examples.
Table 21.1 Sign Conventions for Thin Lenses
Quantity
Convention
s
If object in front* of lens
If object behind lens
If image behind lens
If image in front of lens
If erect
If inverted
If surface is convex
If surface is concave
If converging
If diverging
s
h, h
R1, R2
f
FIGURE 21.4 Geometry to derive
lens equation and magnification
(Equations (21.4) and (21.5)).
s-f
O
F
h
h
C
I
s
s
*
Front and back are with respect to incident light; that is, the front of the lens faces the
incident light.
OPTICAL LENSES
f
f
525
I
F
O
FIGURE 21.5 Raytracing when the
object O is within one focal length
of a converging lens. The image I is
virtual, erect, and magnified.
First consider the situation in Figure 21.5 where the
object is closer to the lens than one focal length. In this case,
Equation (21.4) predicts that s will be negative because
1/s 1/f. What does this mean in terms of the image? The
figure shows raytracing in this case. It is clear that the
F
focused rays do not converge and that therefore there is no
real image, no place at which a screen can be put to see an
image. On the other hand, a viewer on the far side of the
lens from the object looking back through the lens will see
the rays appear to emanate from a (virtual) image behind
the lens, to be larger than the object, and to be erect. That the image is larger and erect
follows from Equation (21.5) because s is greater than s and is negative, making
m 1 (note the agreement with our sign conventions in Table 21.1). As the object
approaches the focal point, the virtual image recedes to larger distances and is magnified to ever greater size. You may recognize this application of a lens as a magnifying
glass (Figure 21.6). We discuss this situation further after we take a look at the eye
in Section 3.
As a second application of the lens equation consider the diverging lens
shown in Figure 21.7. According to the lens-maker equation, because the radii
of curvature are both taken as negative, so is the focal length. Therefore, no matter where an object is placed on the left of the lens in the figure, according to
the lens equation, the image will always be virtual with the object appearing
smaller and erect. This is so because with 1/s 0, when subtracted from 1/| f |,
we have that
1
1
1
s¿
ƒfƒ s
FIGURE 21.6 A happy magnifying
glass.
so that we must always have s 0 and |s| s. Figure 21.7 shows the raytracing diagram for one such situation.
As a final case, we examine the problem of where to put a converging lens in
order to image an object on a screen when the total object to screen distance is fixed
at a distance L. In that case (s s) L, and we must find the possible lens locations, or the possible individual s and s values. Writing the lens equation as
s s¿
1
1
1
, or ss¿ fL,
s s¿
ss¿
f
we need to solve for possible s and s values. Substituting for s (L s)
into the above, we have
s(L s) fL or s2 sL fL 0.
Solving the quadratic equation, there are two possible solutions given
by
s
L ;
1L2 4fL
2
or s 4f
L
L
;
1 ,
2
2A
L
as long as f L/4 or L 4f. To each of these values of s, let’s call them
s and s, both of which are positive, there corresponds a value of
s (s L s) and a magnification m (s/s). Because the two values
s and s add up to L, it is clear that the two solutions are s s and
s s on the one hand and s s and s s on the other. In the first
case, the lens is closer to the screen than to the object and the magnification is less than 1. The image will be inverted and reduced in size and, of
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OPTICAL LENSES
AND
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O
FIGURE 21.7 Raytracing for a
diverging lens to form the image I
of an object O. Ray 1 is parallel
to the optic axis and diverges as
if it originated at the focal point;
ray 2 is aimed at the focal point
on the other side of the lens and
emerges parallel to the axis; ray
3 passes undeviated through the
lens center. The virtual image is
found at the extrapolated location
where these rays cross.
1
2
3
I
F
F
f
f
s
s'
course, real because it is actually formed on a screen. In the second case, the lens
is closer to the object and the real image is enlarged but still inverted (Figure 21.8).
With a handheld magnifying glass this is an easy and interesting experiment to try.
Example 21.2 A 0.2 cm tall object lies 10 cm from a 25 cm focal length magnifying glass. Find the location, magnification, and size of the image. Is it erect or
inverted? Real or virtual?
Solution: By direct substitution into the lens Equation (21.4), using s 10 cm
and f 25 cm, we find that s 16.7 cm. Because s 0, we know that the
image is on the same side of the lens as the object and therefore a virtual image.
The magnification is given by m s/s 1.7, so that the object appears to be
(0.2)(1.7) 0.34 cm tall and erect.
As was mentioned above, many lenses are compound or thick lenses composed
of multiple lenses cemented together, whereas other optical systems may consist of multiple individual thin lenses. Situations in which there are multiple
thin lenses can be handled using the formalism of this section. One
begins by finding the image of the object in the first lens (closest to
object) and then simply treats this image as the object to be imaged
by the second lens, and so on. Using the same consistent sign convention given in Table 21.1, such problems can be analyzed without
any new concepts. For example, if the two (thin) lenses are in contact,
then we can derive a simple formula for the overall focal length of the
combination as follows. Writing the lens equation for the first lens we
have that
FIGURE 21.8 The two solutions to
imaging an object on a screen a
fixed distance away from an object.
1
1
1
,
s1 s1¿
f1
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with a similar equation for the second lens,
1
1
1
.
s2 s2¿
f2
But, using the first image as the object for the second lens means
that s2 s1¿, so that on adding the two equations together we
find that
FIGURE 21.9 Spherical aberration.
Parallel rays far off-axis will focus at
different distances along the optic
axis (horizontal double-headed
arrow). The image in the paraxial
focal plane, shown by the vertical
line, will therefore be blurred laterally (vertical double-headed arrow).
1
1
1
1
.
s1 s2¿
f1 f2
This equation can be interpreted as treating the two lenses in contact with each other
as a single lens with a combined focal length f given by
1
1
1
+
.
f1
f2
f
(21.7)
Most lenses in optical devices are, in fact, compound lenses designed to compensate for aberrations and so Equation (21.7) tells how to find the net focal
length of the compound lens. We use this idea to analyze the compound microscope in the next section. But what is the purpose of cementing multiple lenses
together?
All lenses suffer from various defects in the quality of the image they produce.
Collectively these are termed lens aberrations. We can distinguish two classes of
aberrations: monochromatic, those involving a single color, and chromatic, due to
the dispersion of the lens material, refracting different wavelengths (colors) differently due to a variation in index of refraction. There are five major monochromatic
aberrations, all of which distort the imaging of a single point of the object to a single point of the image. One of these, spherical aberration, is simply due to the
spherical lens curvature giving rise to distortion when incident rays are far from the
optic axis. In particular, as shown in Figure 21.9, parallel rays from a distant point
source arriving at different distances from the optic axis will be imaged at slightly
different points on the axis, resulting in a blurred image of the point object.
Limiting the accepted rays to paraxial rays close to the optic axis with a stop, or
aperture, can reduce spherical aberration. The four other monochromatic aberrations have to do with off-axis imaging and examples of their images are shown in
Figure 21.10.
1. Coma with the comet Hyakutake showing its shape.
COMA
A
B
C
B’
C’
C’
B’
C
B
C
image
object
FIGURE 21.10 Continued
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OPTICAL LENSES
AND
DEVICES
2. Distortion with an example of barrel distortion from a wide-angle camera shot of
a brick wall.
DISTORTION
object
Image
(pincushion
distortion)
Image
(barrel
distortion)
3. Curvature of field with the painting “Anna’s Bedroom” by Scott Kahn illustrating this aberration.
CURVATURE OF FIELD
object
Curved image
plane
Ideal
image
4. Astigmatism with a painting illustrating this aberration.
ASTIGMATISM
object
Imageradial lines
in focus
Imagetangential
lines in
focus
FIGURE 21.10 The four other monochromatic aberrations, other than spherical
aberration, with schematics and example photos.
Chromatic aberrations due to the dispersion of glass result in different focal points
for each color (Figure 21.11). The shorter wavelength light (violet end of the visible
spectrum) experiences a larger index of refraction and is therefore refracted more and
brought to a closer focal point. Compound lenses made from a converging and a diverging lens of different index of refraction glasses are designed to minimize chromatic
aberration and are known as achromatic lenses (Figure 21.12). The longer optical path
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FIGURE 21.11 Chromatic aberration, showing the effect of dispersion on the focusing of different
colored light in a white light beam.
Shorter wavelengths experience
greater refraction due to the higher
index of refraction.
CHROMATIC ABERRATION
of the longer wavelength light in the diverging lens seen in the figure compensates for
the smaller index of refraction at these wavelengths and brings the various colors to a
common focus. All good quality cameras are made with achromatic lenses, the better
ones having very thick multiple lenses to minimize other distortions as well.
2. THE HUMAN EYE
FIGURE 21.12 Using an achromatic lens (compound lens corrected for chromatic aberration)
there is much less chromatic blur.
ACHROMATIC LENS
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Eyes are our visual window to the world. They are complex structures that are capable of very high resolution, are extremely sensitive to light, capable even of detecting single photons, can give color and depth perception, and adjust to focus on
objects from as close as 10 cm, in young eyes, to “infinity.” In this section we first
describe the structure of the eye with the aim of relating its anatomy to its functioning, as well as to some diseases. Then we focus on the retina and the visual pigment
to describe in some detail the actual transduction of photon energy to an electrical
response in the optic nerve.
A schematic cross-section of the human eye is shown in Figure 21.13. Starting
from the outside, the external covering of the eye consists of three layers. Most of the
outermost layer is the sclera, the white of the eye, a tough fibrous layer containing
nerve endings but no blood vessels. The sclera covers about 85% of the eyeball,
roughly a 2.5 cm sphere, but the front portion consists of a transparent 12 mm diameter cornea with an index of refraction of about 1.38. The cornea is the most refracting surface in the eye, with the largest index transition from n 1 in air. The next
two inner layers are the choroid, filled with pigments and blood vessels, and the
retina, the site of photon detection. Neither of these layers extends into the cornea
region (see Figure 21.13).
At the front end of the eye behind the cornea is a liquid-filled chamber with the
aqueous humor, which is continually drained and replaced, bounded also by the
lens and the iris. A buildup of pressure in this region can produce a condition
known as glaucoma, which can lead to blindness. The lens is a double convex lens
made from a crystalline array of 25% protein and 10% lipids, having an index of
refraction of about 1.42. It is one of the few parts of our bodies that
are preserved without any turnover of their cells. With age, or disease, the lens loses its perfect crystallinity and develops defects
that scatter light. These are known as cataracts and, when sufficiently large, can adversely affect vision by “clouding” the eye, or
scattering light just as if you tried to view the world through a thin
layer of milk. The shape of the lens is controlled by ciliary muscles
that can change its focusing ability in a process known as accommodation. Normally, without any shape change of the lens, we can
focus on objects from about 20 feet to infinity. This ability is due
to the finite thickness of the photon detection region allowing light
OPTICAL LENSES
AND
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to be focused at slightly different distances (see below). To see
objects closer than that distance, the eye cannot remain relaxed,
but the lens must change shape, thickening to give a tighter
focus.
The iris serves as an adjustable aperture and is pigmented, giving the eye its characteristic coloring. The central opening, known as
the pupil, is the photon entrance path. Filling the eyeball is a gel-like
material, the vitreous humor, which is more or less permanent. Six
pairs of muscles control the movement of the eyeball in its socket,
allowing us to focus images of interest on the highest acuity region
of the retina, the fovea. This region of the retina, also known as the
macula, has only cones, the photon receptor cells responsible for
color vision, with each cone having a direct connection to a different optic neuron, or nerve cell. The macula therefore has the highest
spatial resolution on the retina; outside the fovea there are roughly
10 cones per neuron connection or 125 rods, the other type of photon receptor, per neuron. These neurons collect in the optic disc,
creating a blind spot with no visual pigment, and lead to the optic
nerve bundle.
Before we consider the photon detection process on the retina in more detail,
consider the overall optical arrangement of the eye. As shown in Figure 21.14, parallel rays of light (from an object at “infinity”) are focused by the relaxed lens to a
point on the retina some 2 cm behind the lens, in fact at the central fovea which lies
on the visual axis. An object at a large but finite distance is focused onto the retina
as an inverted image. Our brain interprets this inverted image as erect. The total
equivalent lens of the eye is a thick lens system, composed of the cornea, aqueous
humor, and lens, subject to all of the aberrations mentioned in the last section. One
function of the iris is to reduce the aperture size to limit incoming rays to be paraxial, thus reducing aberrations.
Often the eyeball is either elongated along the visual axis (myopia, or nearsightedness) or shortened in that direction (hyperopia, or farsightedness). A third
defect in which the cornea is not spherical, but oval in shape, having different focusing properties along two different orthogonal directions, is known as astigmatism.
All three of these defects can be corrected for by placing lenses in front of the eye
(either as eyeglasses or contact lenses). Figure 21.15 shows ray diagrams for each
of these, together with their corrections through the use of a lens. In myopia (top),
parallel rays are brought to a focus in front of the retina (hence the name nearsightedness), blurring the image on the retina. By using a diverging lens, the image
can be formed on the retina. The worse the myopia, the greater the power of the corrective lens needed. In hyperopia (middle), parallel rays would be focused behind
the retina (far-sightedness) and so have not yet converged to form a clear image on
the retina. A converging lens will move the focal point onto the retina. Again, the
worse the eye’s vision is, the stronger power corrective lenses needed. Inexpensive
“reading glasses,” simply matched converging plastic lenses, are sold in various
diopter ratings for several dollars and are usually adequate for reading purposes.
Astigmatism (bottom) produces a distortion in imaging so that two perpendicular
lines cannot be both brought into focus. A cylindrical lens that focuses light along
only one axis can correct this defect.
Example 21.3 Suppose the focal length of a person’s eye is 3.0 cm when fully
relaxed (looking at a distant object). If the person’s retina is 3.3 cm behind the
eye lens (a nearsighted eye compared to the normal distance of 3.0 cm), what
must be the focal length of the corrective lenses so that this person can see
“objects at infinity?”
THE HUMAN EYE
FIGURE 21.13 The human eye
in cross-section.
FIGURE 21.14 A distant object
imaged on the retina.
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Solution: Using the thin lens equation with s and s¿ 3.3 cm, we find an
effective focal length of 3.3 cm needed. Because the effective focal length of
such a two-lens system (the lens of the eye and the corrective lenses) is given,
from Equation (21.7), by
1
feffective
1
flens
1
feye
,
we can find the focal length of the needed lens to be
flens a
1
feffective
1
feye
b
1
a
1
1 1
b 33.0 cm
3.3 3.0
(or in diopters, (1/.33 m) 3 D).
There are some interesting points to be made about nearsightedness. For example, if the near point of the normal eye is 25 cm (smin) and the distance to the retina
is 3 cm (s), then the minimum focal length of the normal eye is 2.7 cm (from
1 1
1
b.
s s¿
f
Normal eyes produce an inverted real image on the retina that is m s¿/s 3 cm/25 cm 0.12 times as large as the object. Let’s assume the nearsighted person
in Example 21.3 has the same minimum focal length. Then, without corrective lenses,
at closest focus
1
1
1
=
+
,
smin
2.7
3.3
FIGURE 21.15 Three common
focusing problems with the eye and
their correction with eyeglasses.
which leads to smin 14.9 cm. Nearsighted people can see clearly closer to their eyes
than normal-sighted persons; they have a smaller near point, which is the closest distance an object can be brought into focus, without corrective glasses. The size of the
real image for this case is 3.3 cm/14.9 cm 0.22, almost twice as large as that of the
normal-sighted person! Nearsighted people who are stamp or coin collectors have a
great advantage over normal-sighted persons; they often don’t need magnifying
glasses to see fine details.
Unfortunately for the nearsighted, this close vision advantage vanishes when
they put on corrective lenses. Suppose the person above is wearing the 3 D corrective lenses designed for distance vision and is looking at an object 25 cm away. The
3 D lens turns out to create a virtual image 14 cm in front of the
lens that serves as the object for the eye’s lens, using
1 1
1 1 1
b 0.14 m.
s¿ a b a 3 s
f
0.25
Near-sighted eye
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Near-sighted eye, corrected
Far-sighted eye
Far-sighted eye, corrected
Astigmatic eye
Astigmatic eye, corrected
But, that is approximately the closest the bare eye lens can focus,
therefore with corrective lenses on, the person can no longer see as
close as without them. (An object 14 cm away would form a virtual image only 10 cm in front of the glasses, too close to be in
focus.) Note also that the virtual image formed by the corrective
lenses is smaller than the object by a factor of 14 cm/25 cm 0.56. This reduction cancels the magnification advantage the nearsighted person enjoyed, too. In fact, because the corrective lens is
diverging, it always forms a reduced size virtual image in front of
OPTICAL LENSES
AND
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the real object. The eye uses this image as its object and so nearsighted people with
corrective lenses always perceive objects to be both closer to them and smaller, compared with what a normal person sees.
As the eye ages, the lens shape becomes more resistant to change by the ciliary muscles and so the eye cannot accommodate as well to bring close objects
into focus. This weakening of accommodation is known as presbyopia. In the young
eye, the near point can be as short as 7 cm. With age, the near point moves farther
away due to presbyopia, having a mean of about 1 m for a 60 year old individual.
Producing a similar effect as hyperopia, this can be corrected with a converging lens.
People with myopia will often see an improvement in their vision with age due to
presbyopia and will often require bifocal lenses with the lower portion made converging for reading or close vision and the upper portion made diverging for distance vision.
The structure of the retina is shown in cross-section in Figure 21.16. Note the
striking fact that incident light must travel through the network of nerve cells (retinal
ganglion cells) before reaching the photoreceptors, lying partly immersed in a layer
of pigmented cells. Fortunately these cells are transparent, but only about 50% of the
light that strikes the cornea gets to the retina, and only about 20% of that gets to the
light detecting cells. These cells, the rods and cones, permanent and not replaced over
time, are, however, 100% efficient. Light that is not absorbed by the photoreceptors
FIGURE 21.16 The human retina. (top) Light passes through a network of nerve cells
from the bottom of this drawing before being detected by the rods and cones. (bottom)
Fluorescence microscopy image of the retina.
THE HUMAN EYE
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FIGURE 21.17 Hexagonally packed
cone cells at the fovea section of
the retina in a living human eye.
The image is taken at a location
about 300 m from the foveal center (which is equal to about
1 degree of visual angle) and it is
about 150 m across. The small
spots are single cone photoreceptors which, at this location are separated by about 5 m. The dark
shadow is that of a blood vessel
which runs above the photoreceptor layer.
FIGURE 21.18 Cone cells of the
human retina.
534
is subsequently absorbed by a layer of pigmented cells to
prevent stray reflections of light within the retina. There are
about 125 million rods and 7 million cones on the retina distributed such that only cones are at the central fovea, where
vision is most acute (Figure 21.17). The rods and cones are
named by their shapes, but are somewhat similar in overall
structure (Figure 21.18).
There is an inner segment, filled with mitochondria
to supply the energy needed for the light transduction,
through which the light must also pass. The retina consumes the greatest amount of oxygen per unit weight of
any tissue in the body. Inner and outer segments are connected through a thin cilialike portion containing microtubules. Light transduction takes place in the outer
segments, the rod outer segments having been studied much
more thoroughly than those of the cone. They are each 20
m long and 2 m in diameter and contain stacks of rod
discs that are membranes containing the visual pigment
rhodopsin, with about 105 rhodopsin molecules per m2 of
surface area.
Rhodopsin consists of two parts: a protein portion with
348 amino acids, known as opsin, and a smaller hydrocarbon part C20H28O, a derivative of vitamin A known as retinal, the light-absorbing
portion (Figure 21.19). The structure of retinal is shown in Figure 21.20. There
are two possible sterioisomers, or conformations, of retinal: 11-cis-retinal found
in the dark and all-trans-retinal that nearly instantaneously forms after the absorption of a single photon. With the advent of femtosecond (1015 s) laser pulses,
this first step in the vision process, the isomerization of retinal, has been found to
occur within about 500 femtoseconds. Subsequent to this initial conformational
change there is a sequence of conformational steps, discovered using pulsed
laser spectroscopy, and other events that lead to an eventual electrical signal at the
neuron. Each photon absorbed by a rhodopsin leads to the hydrolysis of over
100,000 molecules of cyclic GMP, the crucial signaling molecule in the subsequent
transduction. The reduction in cyclic GMP, needed to keep Na channels open
in the rod membrane, causes the eventual polarization of the membrane and electrical signal.
The electrical signal that is sent from the retina to the brain over the optic nerve is
not simply the sum of all rod and cone firings. Somehow the activity of the many rod
and cone interconnections “preprocesses” information about the light falling on them
so that a significant part of “seeing” occurs prior to what goes
on in the visual cortex of the brain. It takes time to perform
the preprocessing of visual information in the rod and cone
networks, perhaps 0.1–0.2 s, a time that matches the response
time of our nervous system.
We do not see the instantaneous values of the electric fields of the EM light waves, which vary about 1015
times per second. Rather we see the effects of electric
fields that have been averaged over many cycles of oscillation. In fact, the signal sent from the eye to the brain is
not directly related to E fields, but rather to the average
intensity, proportional to E2 averaged over many cycles.
Now, rods and cones may detect EM intensity, but not
equally well at all frequencies. The retinal molecule acts
like a damped oscillator. When driven by the oscillating
E field of the light, these molecules vibrate resonantly at
different driving frequencies. There are three kinds of cone
OPTICAL LENSES
AND
DEVICES
FIGURE 21.19 A rod cell on left
with a model of the seven-helix
rhodopsin with the front two
helices of opsin cut away to
show the retinal molecule at the
active site.
cells (with three different resonant frequencies close to “red,” “green,” and “blue”)
and one kind of rod cell, with a resonant peak response between “green” and “blue.”
The absorption spectrum of rhodopsin shown in Figure 21.21 indicates that darkadapted rods are most sensitive (have their strongest absorption) in the green-blue at
500 nm. In strong light this shifts slightly to 550 nm, but with a single absorption
spectrum rods are not able to distinguish all of the different colors we can see in
bright light. Cones are much less sensitive to light (the figure does not show this
because the absorption peaks are all normalized); in fact the cones effectively “turn
off” in dim light. In dim light there is little distinction between colors; everything
appears gray.
In bright light, the cones take over. They have three different types of visual pigments, each having a maximum absorption at a different visible wavelength, corresponding to a different color. It has long been known that (almost) any light color
can be represented as a sum of three “primary” light colors: red (R), green (G), and
blue (B). The RGB system is the basis for color TV, for example. On a color TV
screen, each pixel is divided into an R, a G, and a B subpixel. Three electron beams
sweep rapidly across the screen lighting each pixel with a certain amount of R, of
G, and of B.
Clearly it is tempting to explain the RGB color system in terms of the three
different cone cells. Undoubtedly, there is some connection between the two, but
the connection must be fairly subtle, and, as yet, is still not worked out. One
reason for this situation is that the “R” cone cells actually have their response
maximum at a frequency that is closer to yellow than to red. A second reason is
FIGURE 21.20 The chemical
structure of retinal, the
light-sensitive portion of
rhodopsin.
THE HUMAN EYE
535
FIGURE 21.21 The absorption
spectrum of the “red,” “green,”
and “blue” cones and of the rods
(shown in black).
Absorbance
420
400
498 534 564
500
Wavelength (nm)
600
that some people only have two of the three cone cells, yet many of them seem to
perceive color about as well as people with the normal distribution of cells.
Somehow, their brains fill in the missing information. Finally, in people who are
red–green color blind, all three cells appear to be present. So, the final word is not
in yet.
3. OPTICAL DEVICES: THE MAGNIFYING GLASS
AND OPTICAL MICROSCOPE
FIGURE 21.22 (a) An object at the
near point of the eye subtending an
angle . (b) An object viewed
through a magnifying glass when
placed at its focal point, now
subtending an angle .
h
θ
a
Image at ∞
h
b
536
Optical devices abound in our technologically oriented world, from the supermarket
optical scanner to the sophisticated digital video camera. Nearly all of these devices
incorporate multiple lenses, except for the simple magnifying glass which we study
just below. As we have seen, to analyze imaging problems with multiple lenses we
straightforwardly treat the image from the first lens as the object for the second lens,
and so on. Similarly the overall magnification is the product of the magnifications
from each lens in the combination. We show an example of this in the optical microscope below.
Recall that the near point is the closest distance that an object can be placed
from the eye and remain in focus. If you want to see an object in more detail, you
must bring it even closer to your eyes. In that way the image of the object on the
retina will be enlarged. This allows more detail to be seen, because the image is
then spread out over more detection sites increasing the spatial resolution on the
retina, limited ultimately by the density of nerve cell connections. The unaided
human eye thus has a limited ability to see detail because the near point limits our
ability to bring objects as close as we might like to increase the size of a focused
image on the retina. Figure 21.22a shows a small object at the near point and the
image formed on the retina. If we take the near point to be 25 cm, a typical value,
then the angle subtended by the object is equal to h/25 cm as shown in the
diagram.
To increase the image size even further on the retina, but still have the
image in focus, a magnifying glass (convex lens) is needed. The converging
lens increases the focusing ability of the lens of our eye and allows us to
bring the object closer to our eye while still keeping the image in focus. The
angular magnification, or magnifying power, is defined in terms of the
increased angle subtended by the object as compared to that at the near point
of the unaided eye (see Figure 21.22)
θ
f
mu u¿
.
u
(21.8)
OPTICAL LENSES
AND
DEVICES
Using a magnifying glass with a relaxed eye focused
at infinity and with the eye directly behind the magnifier, a
virtual image at infinity is formed when the object is
placed at the close focal point of the magnifier. Under
these conditions h/f as shown in the diagram and we
can write the angular magnification as
fobj
L
feye
Image at ∞
mu h
a b
f
a
h
b
25 cm
25 cm
.
f
(21.9)
The smaller the focal length of the lens, the greater is the magnification seen
by the eye. The exact position of the object relative to the focal point of the magnifier turns out to be unimportant, only changing the magnification by a small
amount. The eye accommodates these small changes to make the image clear on
the retina resulting in a small increase in magnification. The maximum magnification occurs when the image viewed through the magnifying glass occurs at the
near point of the eye.
To obtain higher magnification still, a compound microscope can be used.
Figure 21.23 shows a schematic drawing of such a microscope with two lenses, an
eyepiece that functions as a magnifying glass and an objective lens that further magnifies the object. The overall magnification is the product of that produced by each
lens. The object is placed just outside the focal point of the objective, s ~ fobj, so that
an inverted real image is formed with a lateral magnification of
mobj FIGURE 21.23 Optics of a simple
compound microscope. The object
is placed just outside the focal
point of the objective lens so that
an enlarged, inverted real image is
formed just inside the focal point of
the eyepiece lens. Its image is a
further enlarged virtual image
viewed at infinity by a relaxed eye.
s¿
s¿
,
s
fobj
from Equation (21.5). This image then acts as the object for the eyepiece, adjusted to
place the final virtual image at infinity, so that the eye can be relaxed as it views
the image. In this case, we can write that s¿ 1L feye2, where L is the distance
between the lenses, as shown in the diagram. The overall magnification compared to
that at the near point with the unaided eye is then
m mobj meye a
L feye
fobj
ba
25 cm
25 cm # L
b L
,
feye
fobj feye
(21.10)
FIGURE 21.24 A basic compound
microscope.
Dual eyepieces
because feye is generally much smaller than L, where all distances are
given in cm. Usually the short focal length lenses of a microscope are
compound lenses designed to eliminate aberrations. Magnifications of
over 1000 are readily obtained.
Figure 21.24 shows a photo of a basic compound microscope with its
three main features: the built-in light source or condenser, providing a
uniform brightness with the use of lenses; a stage, designed to securely
hold the sample and usually to move it about in the horizontal plane; and
the barrel of the microscope, holding the lenses and usually allowing different objectives with different focal lengths to be used. Microscopy has
developed substantially in the recent past. The use of high-sensitivity
video cameras and electronic image-processing techniques, as well as the
development of several new types of microscopy discussed in Chapter 24,
have broadened the versatility of the microscope in studying fundamental
processes in biology.
O P T I C A L D E V I C E S : T H E M AG N I F Y I N G G L A S S
AND
O P T I C A L M I C RO S C O P E
Objective lenses on
rotating mount
x–y moveable stage
Light source
537
CHAPTER SUMMARY
Thin optical lenses have a focal length that is given by
the lens-maker formula
1
1
1
(n 1)a b ,
f
R1 R2
(21.1)
where n is the index of refraction of the lens material
and R1 and R2 are the radii of curvature of the two
faces of the lens. An object located a distance s from
the lens will produce an image through a thin lens
of focal length f at a distance s given by the lens
equation,
If two thin lenses are placed in contact, the overall focal
length of the pair is given by
1
1
1
+
= .
f1
f2
f
The structure, optics, and detection properties of
the eye are discussed in Section 2 of this chapter with
a discussion of some common vision disorders and
their correction with lenses.
A simple magnifying glass of focal length f will
produce a virtual image with a magnification given by
mu 1
1
1
.
s s¿
f
h¿
s¿
.
s
h
(21.9)
where 25 cm is taken as the near point of the eye.
Similarly, the overall magnification of a compound
microscope, made from an eyepiece and an objective
lens, is given by
(21.5)
These three equations work under a wide variety of
conditions if one uses the sign conventions of Table 21.1.
QUESTIONS
1. Distinguish carefully between a converging and
diverging lens, a thin and a thick lens, and a real and
a virtual image. Which combinations do not exist?
For example, is there a thin diverging lens that forms
a real image?
2. Carefully define all the symbols in the lens-maker equation. In a thin double convex lens made from n 1.5
glass with both sides having 25 cm radii of curvature,
what is the predicted focal length?
3. Review the raytracing algorithm for finding the image
of a (real) object through a thin lens. Distinguish the
four cases of a converging lens with object closer or
farther than the focal length, and a diverging lens with
the object closer or farther than a focal length. Make a
sketch of an example of each case. Classify each case
according to whether the image is (erect or inverted),
(real or virtual), (magnified more or less than 1), and
(closer or farther than a focal length from the lens).
4. Consider the object–convex lens–real image configuration. If one places one’s eye at the image location, one
will not see the image; however, if one moves the eye
farther back, away from both the lens and the image
location, the image will become visible. Why is this?
538
25 cm
,
f
(21.4)
The lateral magnification of the image is
m
(21.7)
m mobj meye L
25 cm # L
,
fobj feye
(21.10)
where L is the lens separation distance.
5. Compare the (real) image of a person formed by a converging lens to the (virtual) image of that person in a
plane mirror. In which is up/down reversed, left/right
reversed, and the magnification possibly changed?
6. In applications where a large light source is to be
focused, such as a lighthouse, stage lighting in a theater, or an overhead projector, both a large diameter
and a thick lens are needed. Fresnel, realizing that the
refraction occurs at the glass surface, designed a lens
(now called a Fresnel lens) that keeps the large curvature and lens size, but collapses the lens down to a
nearly planar lens by removing the glass interior. This
overcomes the problem of weight, bulk, and cost of
such a glass lens. Based on the figure below explain
how this lens works.
OPTICAL LENSES
AND
DEVICES
7. What is the physical origin of chromatic aberration in
a thin lens? Of spherical aberration? Which rays are
brought to focus closer to the lens: blue or red? paraxial (those parallel to and close to the optic axis) or offaxis rays?
8. Why does closing down the iris reduce aberrations of
the lens of the eye?
9. One way to correct the eye for myopia is with laser
surgery called photorefractive keratectomy (PRK) in
which the cornea is reshaped so that light from a distance object focuses on the retina instead of in front
of the retina. What do you think the laser procedure
does to the cornea? This surgery will not correct for
the age-related presbyopia that leads to the need for
reading glasses.
10. Discuss the origin of presbyopia and the need for
reading glasses as one ages.
11. Which photoreceptors, rods or cones, give us our
most acute vision? Our color vision? Our night
vision?
12. Explain in basic terms why the magnification of a
microscope (or any two-lens system) is equal to the
product of the magnifications of the objective and
eyepiece.
MULTIPLE CHOICE QUESTIONS
1. A light-emitting object is 10 cm from a thin lens. An
upright virtual image is formed 20 cm behind the
object. Which of the following is true? The lens has a
focal length of (a) 6.7 cm, (b) 15 cm, (c) 15 cm,
(d) 20 cm.
2. The distance from the eye’s lens to the retina for a
given person is 3.0 cm. This person clearly sees an
object 27 cm in front of his eye. The focal length of
the eye’s lens in this case is (a) 2.7 cm, (b) 3.0 cm,
(c) 3.4 cm, (d) 3.4 cm.
3. A 10 cm tall object 25 cm from a converging lens has
its real image 50 cm from the lens. The object appears
to be (a) 20 cm tall and erect, (b) 5 cm tall and
inverted, (c) 5 cm tall and erect, (d) 20 cm tall and
inverted.
4. A light source with an arrow pointing up is placed at
the zero mark on an optical bench. A convex lens of
unknown focal length is placed with its center at the
30 cm mark on the bench. A focused image appears
on a collector when placed at the 60 cm mark on the
bench and nowhere else. What must be true about the
image? It is (a) real and inverted, (b) real and upright,
(c) virtual and inverted, (d) virtual and upright.
5. The focal length of the lens in the previous question must be (a) 15 cm, (b) 15 cm, (c) 30 cm,
(d) 60 cm.
6. Suppose a concave lens is inserted at the 15 cm mark
on the bench in question 4. What would you have
to do to the collector to find a focused image now?
(a) Leave it at 60 cm. (b) Move it to some position
QU E S T I O N S / P RO B L E M S
7.
8.
9.
10.
11.
12.
13.
14.
15.
between the 30 cm mark and the 60 cm mark. (c) Move
it to a position farther out than the 60 cm mark. (d) You
can’t move the collector anywhere to get a focused
image because no real image will be formed by this
arrangement of lenses.
In drawing a ray diagram for a converging lens with
an object farther away than a focal length which of
the following is not a correct ray to draw: (a) a ray
parallel to the optic axis deflects at the lens to appear
to come from the near focal point, (b) a ray parallel to
the optic axis deflects at the lens to go through the
focal point on the far side of the lens, (c) a ray going
through the near focal point deflects at the lens and
emerges from the lens parallel to the optic axis, (d) a
ray straight through the lens center.
A plano-convex lens of crown glass with a radius
of curvature of 50 cm has a focal length of (a) 1.0 m,
(b) 0.5 m, (c) 2.0 m, (d) 1.0 m.
A double concave lens of crown glass with radii
of curvature magnitudes of 25 cm and 50 cm has a
focal length of (a) 0.33 m, (b) 3 m, (c) 0.33 m,
(d) 33 m.
Chromatic aberration in lenses is due to (a) dispersion, (b) interference, (c) total internal reflection, (d)
varying degrees of absorption of different colors of
light.
Which of the following best describes nearsightedness? The lens in the eye produces an image of an
object (a) 5 m away that would form behind the retina,
(b) 5 m away that forms in front of the retina, (c) 10
cm away that would form behind the retina, (d) 10 cm
away that forms in front of the retina.
Without corrective lenses a woman can see an object
clearly no closer than 0.5 m from her face. With corrective lenses she can see the object clearly as close
as 0.1 m from her face. When the object is at 0.1 m
her corrective lenses must form a (a) real image 0.5 m
in front of her face, (b) real image 0.1 m in front of
her face, (c) virtual image 0.5 m in front of her face,
(d) virtual image 0.1 m in front of her face.
A prescription for corrective lenses reads 5 D for
each lens. These corrective lenses are (a) diverging
with focal length equal to 5 m, (b) diverging with
focal length equal to 0.2 m, (c) converging with focal
length equal to 5 m, (d) converging with focal length
equal to 0.2 m.
It is essentially impossible for humans to react to a
visual stimulus in less than 0.1 s. This is because that
is about how long it takes (a) light to pass from the
lens of the eye to the retina, (b) light to make one
complete cycle of oscillation, (c) molecules in the
cones to complete one cycle of vibration when they
are excited by light, (d) firing from the retinal cells to
be processed before being sent to the brain.
In very dim light you can see an object better by looking at it out of the corner of your eye than straight on.
That is because (a) cones are more concentrated on
539
the lens x, in order that the camera be able to take
sharp photographs of objects positioned anywhere
from 50 cm to infinity, measured from the front surface of the camera body.
the periphery of the retina than rods and cones function better in dim light, (b) cones are more concentrated on the periphery of the retina than rods and
rods function better in dim light, (c) rods are more
concentrated on the periphery of the retina than cones
and cones function better in dim light, (d) rods are
more concentrated on the periphery of the retina than
cones and rods function better in dim light.
16. Color vision is due to (a) the varying sensitivity of
different rhodopsins to different wavelengths of light,
(b) three different kinds of rod cells, (c) the dispersion of the lens of the eye, (d) the pigmented epithelial cells.
17. A magnifying glass produces an image that is (a)
upright and real, (b) inverted and real, (c) upright and
virtual, (d) inverted and virtual.
18. Light entering the eye passes through the following
layers in the order (a) lens, cornea, ganglion cells,
retina, (b) cornea, lens, ganglion cells, retina, (c) lens,
cornea, retina, ganglion cells, (d) cornea, lens, retina,
ganglion cells.
PROBLEMS
1. Sunlight can be focused on the ground by a lens when
it is held 24 cm above the surface. What is the power
of the lens?
2. A 5.0 diopter lens is used to magnify an insect
when held 12 cm away. Describe the type of image,
its position, and lateral magnification. Draw a ray diagram sketch.
3. A pinhole can function as a “lens”. Consider a box
with a very small hole in one side. The hole admits
light from any and all points outside to the inside but
from any one point outside only the ray directed at the
hole can enter the box. Show with a diagram the
image of the object obtained by the light admitted by
the pinhole. What is the magnification? Such boxes
can be used as cameras, provided they are mounted
on a rock-solid surface. Film exposures are typically
many seconds or minutes and the image quality can
be superb.
4. An old-fashioned box camera has a fixed lens and a
depth of 15 cm. Suppose the camera lens is designed
to be optimal for taking photographs of objects 3 m
away. What is the focal length of the lens?
camera
cheese
3m
15 cm
5. A camera has a lens with adjustable position. The
camera depth d 4 cm. Determine the focal length
of the lens and the necessary allowable extension of
540
d
Subject at
minimum
object distance
x
50 cm
6. How far from its infinity setting must a 35 mm lens
be moved so that it produces a sharp image of an
object 3 m away?
7. Suppose the image of a creature swimming in a dish
of water is projected onto a screen. The distance from
the dish to the lens is 36 cm and the screen is 4.5 m
away from the mirror and lens. (The mirror simply
redirects the light from the lens to the screen. Treat
the distance between the lens and mirror as small
enough so that it can be neglected in the specification
of the image distance. Thus, object distance sv is 36 cm
and image distance sh is 4.5 m.)
sv
sh
(a) What is the focal length of the projection lens?
(b) If the creature swims at 1 cm per second in the
dish, how fast does the image move on the screen?
8. A 35 mm film slide projector has a projection lens
with a focal length of 135 mm.
(a) Where should the slide be placed if the projection
screen is 3 m away from the projection lens?
(b) What is the magnification?
(c) Now for the hard part (judging by the difficulty
that even well-educated lecturers have with this
one in practice). If we want to get a true (upright,
nonreversed) image of the slide on the screen,
how should the slide be placed into the projector?
Should it be flipped upside down? Should it be
reversed left and right? Should the slide be
inserted backwards? (What does “backwards”
mean here?)
9. A quick and easy way to get an approximate determination of the focal length of a convex lens is to measure the distance from the lens to an image of a light
or other bright source some distance away. Suppose
one has a lens that in fact has a focal length of 10 cm.
OPTICAL LENSES
AND
DEVICES
A fluorescent ceiling light with a grill cover is 1.5 m
above the lens and a student is able to see an image of
the lighting fixture grill on the back of his hand. He
declares that the focal length of the lens is equal to the
distance between lens and hand. Calculate the actual
hand–lens separation distance for a sharp image and
show that the error in the value of the focal length
determined this way is less than 10%. Error [(focal
length distance measured)/focal length] 100%.
10. Consider a convex lens of focal length 20 cm.
Calculate the image distance for each of the following object distances: , 4 m, 2 m, 1 m, 80 cm, 60 cm,
40 cm, 20 cm.
11. It is often necessary to convey or relay the image of
an object while keeping the image size unchanged
QU E S T I O N S / P RO B L E M S
(unit magnification). Consider an object for which an
image of the same size is desired, exactly 1 m away.
(a) What is the focal length and the location of the
single lens that will accomplish this?
(b) The image of the object, using one lens, will be
inverted. A relay system using two identical convex lenses will invert the inversion, yielding an
upright image. Design such a system for the same
initial situation as in the previous part.
12. If a certain microscope has an eyepiece with 12
magnification and it is desired to view a specimen
with an overall magnification of 60, what is the
power of the objective that must be used?
13. The magnification of a compound microscope can be
slightly improved if the final image is not at infinity
but rather at the near point of the eye. Derive the formula for the magnification under this condition. This
arrangement tends to produce eye strain because the
iris must be under tension to have the lens of the eye
constantly focusing at the near point.
14. Tom Cruise catches a reporter shooting pictures of his
daughter at his home. He claims the reporter was trespassing. To prove his point, he gives as evidence the
film the police took from the reporter. His daughter’s
height of 0.62 m is 2.89 mm high on the film, and the
focal length of the camera lens that the police seized
was 210 mm. How far away from the baby was the
reporter standing? Could the reporter be trespassing?
541