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Unformatted text preview: 27 Geometrical 0ptics 27—1 Introduction In this chapter we shall discuss some elementary applications of the ideas of
the previous chapter to a number of practical devices, using the approximation
called geometrical optics. This is a most useful approximation in the practical
design of many optical systems and instruments. Geometrical optics is either
very simple or else it is very complicated. By that we mean that we can either study
it only superﬁcially, so that we can design instruments roughly, using rules that
are so simple that we hardly need deal with them here at all, since they are practi
cally of high school level, or else, if we want to know about the small errors in
lenses and similar details, the subject gets so complicated that it is too advanced
to discuss here! If one has an actual, detailed problem in lens design, including
analysis of aberrations, then he is advised to read about the subject or else simply
to trace the rays through the various surfaces (which is what the book tells how to
do), using the law of refraction from one side to the other, and to ﬁnd out where
they come out and see if they form a satisfactory image. People have said that this
is too tedious, but today, with computing machines, it is the right way to do it.
One can set up the problem and make the calculation for one ray after another
very easily. So the subject is really ultimately quite simple, and involves no new
principles. Furthermore, it turns out that the rules of either elementary or advanced
optics are seldom characteristic of other ﬁelds, so that there is no special reason
to follow the subject very far, with one important exception. The most advanced and abstract theory of geometrical optics was worked
out by Hamilton, and it turns out that this has very important applications in
mechanics. It is actually even more important in mechanics than it is in optics,
and so we leave Hamilton’s theory for the subject of advanced analytical mechanics,
which is studied in the senior year or in graduate school. So, appreciating that
geometrical optics contributes very little, except for its own sake, we now go on to
discuss the elementary properties of simple optical systems on the basis of the
principles outlined in the last chapter. In order to go on, we must have one geometrical formula, which is the follow
ing: if we have a triangle with a small altitude h and a long base d, then the diagonal
s (we are going to need it to ﬁnd the difference in time between two different routes)
is longer than the base (Fig. 27—1). How much longer? The difference A = s — d
can be found in a number of ways. One way is this. We see that s2 — d2 = W,
or (s — d)(s + d) = W. But .9 — d = A, ands + d ~ 25. Thus A ~ h2/2.r. (27.1) This is all the geometry we need to discuss the formation of images by curved
surfaces! 27—2 The focal length of a spherical surface The ﬁrst and simplest situation to discuss is a single refracting surface, sep
arating two media with diﬁerent indices of refraction (Fig. 27—2). We leave the
case of arbitrary indices of refraction to the student, because ideas are always the
most important thing, not the speciﬁc situation, and the problem is easy enough
to do in any case. So we shall suppose that, on the left, the speed is l and on the
right it is 1/11, where n is the index of refraction. The light travels more slowly in
the glass by a factor n. 27—1 27—1 Introduction 27—2 The focal length of a spherical
surface 27—3 The focal length of a lens
27—4 Magniﬁcation
27—5 Compound lenses 27—6 Aberrations
27—7 Resolving power Figure 27—1 Fig. 27—2. Focusing by a single re
fracting surface. Now suppose that we have a point at 0, at a distance 3 from the front surface
of the glass, and another point 0’ at a distance s’ inside the glass, and we desire to
arrange the curved surface in such a manner that every ray from 0 which hits the
surface, at any point P, will be bent so as to proceed toward the point 0’. For that
to be true, we have to shape the surface in such a way that the time it takes for
the light to go from 0 to P, that is, the distance 0P divided by the speed of light
(the speed here is unity), plus n ' 0’P, which is the time it takes to go from P to 0’,
is equal to a constant independent of the point P. This condition supplies us with
an equation for determining the surface. The answer is that the surface is a very
complicated fourthdegree curve, and the student may entertain himself by trying
to calculate it by analytic geometry. It is simpler to try a special case that corre
sponds to s —> 00, because then the curve is a seconddegree curve and is more
recognizable. It is interesting to compare this curve with the parabolic curve we
found for a focusing mirror when the light is coming from inﬁnity. So the proper surface cannot easily be made—to focus the light from one
point to another requires a rather complicated surface. It turns out in practice
that we do not try to make such complicated surfaces ordinarily, but instead
we make a compromise. Instead of trying to get all the rays to come to a focus, we
arrange it so that only the rays fairly close to the axis 00’ come to a focus. The
farther ones may deviate if they want to, unfortunately, because the ideal surface
is complicated, and we use instead a spherical surface with the right curvature at
the axis. It is so much easier to fabricate a sphere than other surfaces that it is
proﬁtable for us to ﬁnd out what happens to rays striking a spherical surface,
supposing that only the rays near the axis are going to be focused perfectly.
Those rays which are near the axis are sometimes called paraxial rays, and what
we are analyzing are the conditions for the focusing of paraxial rays. We shall
discuss later the errors that are introduced by the fact that all rays are not always
close to the axis. Thus, supposing P is close to the axis, we drop a perpendicular PQ such that
the height PQ is h. For a moment, we imagine that the surface is a plane passing
through P. In that case, the time needed to go from 0 to P would exceed the time
from 0 to Q, and also, the time from P to 0’ would exceed the time from Q to 0’.
But that is why the glass must be curved, because the total excess time must be
compensated by the delay in passing from V to Q! Now the excess time along
route OP is hz/Zs, and the excess time on the other route is nh 2/2s’. This excess
time, which must be matched by the delay in going along VQ, differs from what
it would have been in a vacuum, because there is a medium present. In other
words, the time to go from V to Q is not as if it were straight in the air, but it is
slower by the factor n, so that the excess delay in this distance is then (n — 1)VQ.
And now, how large is VQ? If the point C is the center of the sphere and if
its radius is R, we see by the same formula that the distance VQ is equal to
h2/2R. Therefore we discover that the law that connects the distances 5 and s’,
and that gives us the radius of curvature R of the surface that we need, is (112/25) + (nh2/2s’) = (n — l)h2/2R (27.2)
or (US) + (n/S’) = (n  1)/R (273) If we have a position 0 and another position 0’, and want to focus light from 0
to 0’, then we can calculate the required radius of curvature R of the surface by
this formula. Now it turns out, interestingly, that the same lens, with the same curvature
R, will focus for other distances, namely, for any pair of distances such that the
sum of the two reciprocals, one multiplied by n, is a constant. Thus a given lens
will (so long as we limit ourselves to paraxial rays) focus not only from 0 to 0’,
but between an inﬁnite number of other pairs of points, so long as those pairs of
points bear the relationship that 1 /s + n/s’ is a constant, characteristic of the lens. In particular, an interesting case is that in which 5 —> 00. We can see from the
formula that as one s increases, the other decreases. In other words, if point 0 27—2 goes out, point 0’ comes in, and vice versa. As point 0 goes toward inﬁnity, point
0’ keeps moving in until it reaches a certain distance, called the focal length f’,
inside the material. If parallel rays come in, they will meet the axis at a distance
f’. Likewise, we could imagine it the other way. (Remember the reciprocity rule:
if light will go from 0 to 0’, of course it will also go from 0’ to 0.) Therefore, if
we had a light source inside the glass, we might want to know where the focus is.
In particular, if the light in the glass were at inﬁnity (same problem) where would
it come to a focus outside? This distance is called f. Of course, we can also put
it the other way. If we had a light source at f and the light went through the surface,
then it would go out as a parallel beam. We can easily ﬁnd out what f and f’ are: n/f’ =01 —1)/R or f’ = Rn/(n — 1),
l/f = (n — l)/R or f = R/(n — 1). We see an interesting thing: if we divide each focal length by the corresponding
index of refraction we get the same result! This theorem, in fact, is general. It is
true of any system of lenses, no matter how complicated, so it is interesting to
remember. We did not prove here that it is general—we merely noted it for a single
surface, but it happens to be true in general that the two focal lengths of a system
are related in this way. Sometimes Eq. (27.3) is written in the form Us + n/s’ = l/f. This is more useful than (27.3) because we can measure f more easily than we can
measure the curvature and index of refraction of the lens: if we are not interested
in designing a lens or in knowing how it got that way, but simply lift it off a shelf,
the interesting quantity is f, not the n and the l and the R! Now an interesting situation occurs if s becomes less than f. What happens
then? If s < f, then (Us) > (l/j), and therefore 3’ is negative; our equation
says that the light will focus only with a negative value of s’, whatever that means!
It does mean something very interesting and very deﬁnite. It is still a useful formula,
in other words, even when the numbers are negative. What it means is shown in
Fig. 27—3. If we draw the rays which are diverging from 0, they will be bent, it is
true, at the surface, and they will not come to a focus, because 0 is so close in that
they are “beyond parallel.” However, they diverge as if they had come from a
point 0’ outside the glass. This is an apparent image, sometimes called a virtual
image. The image 0’ in Fig. 27—2 is called a real image. If the light really comes to
a point, it is a real image. But if the light appears to be coming from a point, a
ﬁctitious point different from the original point, it is a virtual image. So when
5’ comes out negative, it means that 0’ is on the other side of the surface, and every
thing is all right. Now consider the interesting case where R is equal to inﬁnity; then we have
(Us) + (n/s’) = 0. In other words, s’ = —ns, which means that if we look from
a dense medium into a rare medium and see a point in the rare medium, it appears
to be deeper by a factor n. Likewise, we can use the same equation backwards,
so that if we look into a plane surface at an object that is at a certain distance
inside the dense medium, it will appear as though the light is coming from not as
far back (Fig. 27—4). When we look at the bottom of a swimming pool from above,
it does not look as deep as it really is, by a factor 3/4, which is the reciprocal of the
index of refraction of water. We could go on, of course, to discuss the spherical mirror. But if one appreci
ates the ideas involved, he should be able to work it out for himself. Therefore
we leave it to the student to work out the formula for the spherical mirror, but
we mention that it is well to adopt certain conventions concerning the distances
involved: (27.4)
(27.5) (27.6) (1) The object distance 3 is positive if the point 0 is to the left of the surface.
(2) The image distance 3’ is positive if the point 0’ is to the right of the surface. (3) The radius of curvature of the surface is positive if the center is to the right
of the surface. 27—3 Fig. 27—3. A virtual image. Fig. 27—4. A plane surface reimages
the light from O’ to O. Fig. 27—5. Image formation by a
twosurface lens. Fig. 27—6. A thin lens with two posi
tive radii. In Fig. 27—2, for example, s, s’, and R are all positive; in Fig. 27—3, s and R are
positive, buts’ is negative. If we had used a concave surface, our formula (27.3)
would still give the correct result if we merely make R a negative quantity. In working out the corresponding formula for a mirror, using the above
conventions, you will ﬁnd that if you put 71 = —1 throughout the formula (27.3)
(as though the material behind the mirror had an index — l), the right formula for
a mirror results! Although the derivation of formula (27.3) is simple and elegant, using least
time, one can of course work out the same formula using Snell’s law, remembering
that the angles are so small that the sines of angles can be replaced by the angles
themselves. 27—3 The focal length of a lens Now we go on to consider another situation, a very practical one. Most of
the lenses that we use have two surfaces, not just one. How does this affect
matters? Suppose that we have two surfaces of different curvature, with glass
ﬁlling the space between them (Fig. 27—5). We want to study the problem of
focusing from a point 0 to an alternate point 0’. How can we do that? The answer
is this: First, use formula (27.3) for the ﬁrst surface, forgetting about the second
surface. This will tell us that the light which was diverging from 0 will appear
to be converging or diverging, depending on the sign, from some other point,
say 0’. Now we consider a new problem. We have a different surface, between
glass and air, in which rays are converging toward a certain point 0’. Where will
they actually converge? We use the same formula again! We ﬁnd that they con
verge at 0”. Thus, if necessary, we can go through 75 surfaces by just using the
same formula in succession, from one to the next! There are some rather highclass formulas that would save us considerable
energy in the few times in our lives that we might have to chase the light through
ﬁve surfaces, but it is easier just to chase it through ﬁve surfaces when the problem
arises than it is to memorize a lot of formulas, because it may be we will never have
to chase it through any surfaces at all! In any case, the principle is that when we go through one surface we ﬁnd a
new position, a new focal point, and then take that point as the starting point for
the next surface, and so on. In order to actually do this, since on the second surface
we are going from n to 1 rather than from 1 to n, and since in many systems there
is more than one kind of glass, so that there are indices n1, n2, . . . , we really need
a generalization of formula (27.3) for a case where there are two different indices, 711 and n2, rather than only M. Then it is not difﬁcult to prove that the general
form of (27.3) is ("1/5) + ("2/3’) = ("2 — n1)/R Particularly simple is the special case in which the two surfaces are very close
together—so close that we may ignore small errors due to the thickness. If we draw
the lens as shown in Fig. 27—6, we may ask this question: How must the lens be
built so as to focus light from 0 to 0’? Suppose the light comes exactly to
the edge of the lens, at point P. Then the excess time in going from 0 to 0’ is
(n1h2/2s) + (nlhz/Zs’), ignoring for a moment the presence of the thickness T
of glass of index n2. Now, to make the time for the direct path equal to that for
the path OPO’, we have to use a piece of glass whose thickness Tat the center is such
that the delay introduced in going through this thickness is enough to compensate
for the excess time above. Therefore the thickness of the lens at the center must
be given by the relationship (nlhz/Zs) + (nlhz/Zs’) = (712 — n1)T. (27.7) (27.8) We can also express T in terms of the radii R1 and R2 of the two surfaces. Paying
attention to our convention (3), we thus ﬁnd, for R1 < R2 (a convex lens), T = (712/212,) — (h2/2R2). (27.9) 274 Therefore, we ﬁnally get ("i/S)+("1/S')= (n2 — "DU/R1 " 1/R2) Now we note again that if one of the points is at inﬁnity, the other will be at a
point which we will call the focal length f. The focal length f is given by 1/f= (n — 1)(1/R1  1/R2), (27.10) (27.11)
wheren = n2/n1. Now, if we take the opposite case, where s goes to inﬁnity, we see that s’ is at
the focal length 1’. This time the focal lengths are equal. (This is another special
case of the general rule that the ratio of the two focal lengths is the ratio of the
indices of refraction in the two media in which the rays focus. In this particular
optical system, the initial and ﬁnal indices are the same. so the two focal lengths
are equal.) Forgetting for a moment about the actual formula for the focal length, if we
bought a lens that somebody designed with certain radii of curvature and a certain
index, we could measure the focal length, say, by seeing where a point at inﬁnity
focuses. Once we had the focal length, it would be better to write our equation in
terms of the focal length directly, and the formula then is (1/3) + (1/S’) = 1/f Now let us see how the formula works and what it implies in different circum
stances. First, it implies that if s or s’ is inﬁnite the other one is f. That means that
parallel light focuses at a distance f, and this in effect deﬁnes f. Another interesting
thing it says is that both points move in the same direction. If one moves to the
right, the other does also. Another thing it says is that s and s’ are equal if they are
both equal to 2f. In other words, if we want a symmetrical situation, we ﬁnd that
they will both focus at a distance 2f. (27.12) 27—4 Magniﬁcation So far we have discussed the focusing action only for points on the axis. Now
let us discuss also the imaging of objects not exactly on the axis, but a little bit off,
so that we can understand the properties of magniﬁcation. When we set up a lens
so as to focus light from a small ﬁlament onto a “point” on a screen, we notice
that on the screen we get a “picture” of the same ﬁlament, except of a larger or
smaller size than the true ﬁlament. This must mean that the light comes to a focus
from each point of the ﬁlament. In order to understand this a little better, let us
analyze the thin lens system shown schematically in Fig. 27—7. We know the follow
ing facts: (1) Any ray that comes in parallel on one side proceeds toward a certain par
ticular point called the focus on the other side, at a distanceffrom the lens. (2) Any ray that arrives at the lens from the focus on one side comes out parallel
to the axis on the other side. This is all we need to establish formula (27.12) by geometry, as follows: Suppose
we have an object at some distance x from the focus; let the height of the object
be y. Then we know that one of the rays, namely PQ, will be bent so as to pass
through the focus R on the other side. Now if the lens will focus point P at all, we
can ﬁnd out where if we ﬁnd out where just one other ray goes, because the new
focus will be where the two intersect again. We need only use our ingenuity to
ﬁnd the exact direction of one other ray. But we remember that a parallel ray goes
through the focus and vice versa: a ray which goes through the focus will come out
parallel! So we draw ray PT through U. (It is true that the actual rays which are
doing the focusing may be much more limited than the two we have drawn, but
they are harder to ﬁgure, so we make believe that we can make this ray.) Since it
would come out parallel, we draw TS parallel to X W. The intersection S is the
point we need. This will determine the correct place and the correct height. Let 27—5 Fig. 27—7. The geometry of imaging
by a thin lens. Fig. 27—8.
planes of an optical system. Illustration of the principal us call the height y’ and the distance from the focus, x'. Now we may derive a
lens formula. Using the similar triangles PV U and TX U, we ﬁnd 1” y
— = —' 2 .l
f x ( 7 3)
Similarly, from triangles S WR and QXR, we get
y’ _ z.
x, — f (27.14)
Solving each for y’/ y, we ﬁnd that
y’ x’ f
— = — = — 27.15
y f x ( ) Equation (27.15) is the famous lens formula; in it is everything we need to know
about lenses: It tells us the magniﬁcation, y’/ y, in terms of the distances and the
focal lengths. It also connects the two distances x and x’ with f: xx’ = f2, (27.16)
which is a much neater form to work with than Eq. (27.12). We leave it to the student to demonstrate that if we call s = x + f and s’ = x’ + f, Eq. (27.12)
is the same as Eq. (27.16). 27—5 Compound lenses Without actually deriving it, we shall brieﬂy describe the general result when
we have a number of lenses. If we have a system of several lenses, how can we
possibly analyze it? That is easy. We start with some object and calculate where
its image is for the ﬁrst lens, using formula (27.16) or (27.12) or any other equivalent
formula, or by drawing diagrams. So we ﬁnd an image. Then we treat this image
as the source for the next lens, and use the second lens with whatever its focal
length is to again ﬁnd an image. We simply chase the thing through the succession
oflenses. That is all there is to it. It involves nothing new in principle, so we shall
not go into it. However, there is a very interesting net result of the effects of any
sequence of lenses on light that starts and ends up in the same medium, say air.
Any optical instrument~a telescope or a microscope with any number of lenses
and mirrors—has the following property: There exist two planes, called the
principal planes of the system (these planes are often fairly close to the ﬁrst surface
of the ﬁrst lens and the last surface of the last lens), which have the following prop
erties: (1) If light comes into the system parallel from the ﬁrst side, it comes out
at a certain focus, at a distance from the second principal plane equal to the focal
length, just as though the system were a thin lens situated at this plane. (2) If
parallel light comes in the other way, it comes to a focus at the same distance f
from the ﬁrst principal plane, again as if a thin lens where situated there. (See
Fig. 27—8.) Of course, if we measure the distances x and x’, and y and y’ as before, the
formula (27.16) that we have written for the thin lens is absolutely general, pro
vided that we measure the focal length from the principal planes and not from the
center of the lens. It so happens that for a thin lens the principal planes are coin
cident. It is just as though we could take a thin lens, slice it down the middle, and
separate it, and not notice that it was separated. Every ray that comes in pops out
immediately on the other side of the second plane from the same point as it went
into the ﬁrst plane! The principal planes and the focal length may be found either
by experiment or by calculation, and then the whole set of properties of the optical
system are described. It is very interesting that the result is not complicated when
we are all ﬁnished with such a big, complicated optical system. 27—6 27—6 Aberrations Before we get too excited about how marvelous lenses are, we must hasten
to add that there are also serious limitations, because of the fact that we have
limited ourselves, strictly speaking, to paraxial rays, the rays near the axis. A real
lens having a ﬁnite size will, in general, exhibit aberrations. For example, a ray
that is on the axis, of course, goes through the focus; a ray that is very close to the
axis will still come to the focus very well. But as we go farther out, the ray begins
to deviate from the focus, perhaps by falling short, and a ray striking near the
top edge comes down and misses the focus by quite a wide margin. So, instead of
getting a point image, we get a smear. This effect is called spherical aberration,
because it is a property of the spherical surfaces we use in place of the right shape.
This could be remedied, for any speciﬁc object distance, by reforming the shape of
the lens surface, or perhaps by using several lenses arranged so that the aberrations
of the individual lenses tend to cancel each other. Lenses have another fault: light of different colors has different speeds, or
different indices of refraction, in the glass, and therefore the focal length of a
given lens is different for different colors. So if we image a white spot, the image
will have colors, because when we focus for the red, the blue is out of focus, or
vice versa. This property is called chromatic aberration. There are still other faults. If the object is off the axis, then the focus really
isn‘t perfect any more, when it gets far enough off the axis. The easiest way to
verify this is to focus a lens and then tilt it so that the rays are coming in at a large
angle from the axis. Then the image that is formed will usually be quite crude,
and there may be no place where it focuses well. There are thus several kinds of
errors in lenses that the optical designer tries to remedy by using many lenses to
compensate each other’s errors. How careful do we have to be to eliminate aberrations? Is it possible to make
an absolutely perfect optical system? Suppose we had built an optical system that
is supposed to bring light exactly to a point. Now, arguing from the point of view
of least time, can we ﬁnd a condition on how perfect the system has to be? The
system will have some kind of an entrance opening for the light. If we take the
farthest ray from the axis that can come to the focus (if the system is perfect, of
course), the times for all rays are exactly equal. But nothing is perfect, so the
question is, how wrong can the time be for this ray and not be worth correcting
any further? That depends on how perfect we want to make the image. But
suppose we want to make the image as perfect as it possibly can be made. Then,
of course, our impression is that we have to arrange that every ray takes as nearly
the same time as possible. But it turns out that this is not true, that beyond a
certain point we are trying to do something that is too ﬁne, because the theory of
geometrical optics does not work! Remember that the principle of least time is not an accurate formulation,
unlike the principle of conservation of energy or the principle of conservation of
momentum. The principle of least time is only an approximation, and it is inter
esting to know how much error can be allowed and still not make any apparent
difference. The answer is that if we have arranged that between the maximal ray—
the worst ray, the ray that is farthest out—and the central ray, the difference in
time is less than about the period that corresponds to one oscillation of the light,
then there is no use improving it any further. Light is an oscillatory thing with a
deﬁnite frequency that is related to the wavelength, and if we have arranged that
the time difference for different rays is less than about a period, there is no use
going any further. 27—7 Resolving power Another interesting question—a very important technical question with all
optical instruments—is how much resolving power they have. If we build a micro—
scope, we want to see the objects that we are looking at. That means, for instance,
that if we are looking at a bacterium with a spot on each end, we want to see that 277 Fig. 279.
optical system. The resolving power of on there are two dots when we magnify them. One might think that all we have to
do is to get enough magniﬁcation—we can always add another lens, and we can
always magnify again and again, and with the cleverness of designers, all the
spherical aberrations and chromatic aberrations can be cancelled out, and there
is no reason why we cannot keep on magnifying the image. So the limitations ofa
microscope are not that it is impossible to build a lens that magniﬁes more than
2000 diameters. We can build a system of lenses that magniﬁes 10,000 diameters,
but we still could not see two points that are too close together because of the
limitations of geometrical optics, because of the fact that least time is not precise. To discover the rule that determines how far apart two points have to be so
that at the image they appear as separate points can be stated in a very beautiful
way associated with the time it takes for different rays. Suppose that we disregard
the aberrations now, and imagine that for a particular point P (Fig. 27—9) all the
rays from object to image T take exactly the same time. (It is not true, because it
is not a perfect system, but that is another problem.) Now take another nearby
point, P’, and ask whether its image will be distinct from T. In other words, whether
we can make out the difference between them. Of course, according to geometrical
optics, there should be two point images, but what we see may be rather smeared
and we may not be able to make out that there are two points. The condition that
the second point is focused in a distinctly different place from the ﬁrst one is that
the two times for the extreme rays PST and P’RT on each side of the big opening
of the lenses to go from one end to the other, must not be equal from the two
possible object points to a given image point. Why? Because, if the times were
equal, of course both wouldfocus at the same point. So the times are not going to
be equal. But by how much do they have to differ so that we can say that both do
not come to a common focus, so that we can distinguish the two image points?
The general rule for the resolution of any optical instrument is this: two different
point sources can be resolved only if one source is focused at such a point that
the times for the maximal rays from the other source to reach that point, as
compared with its own true image point, differ by more than one period. It is
necessary that the difference in time between the top ray and the bottom ray to
the wrong focus shall exceed a certain amount, namely, approximately the period
of oscillation of the light: 12 — 11 > l/u, (27.17) where 1/ is the frequency of the light (number of oscillations per second; also speed
divided by wavelength). If the distance of separation of the two points is called
D, and if the opening angle of the lens is called 0, then one can demonstrate that
(27.17) is exactly equivalent to the statement that D must exceed A/n sin 0, where
n is the index of refraction at P and A is the wavelength. The smallest things that
we can see are therefore approximately the wavelength of light. A corresponding formula exists for telescopes, which tells us the smallest difference in angle between
two stars that can just be distinguished.* * The angle is about )\/ D, where D is the lens diameter. Can you see why? 27—8 ...
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This note was uploaded on 06/18/2009 for the course PHYSICS none taught by Professor Leekinohara during the Spring '09 term at Uni. Nottingham  Malaysia.
 Spring '09
 LeeKinohara
 Physics

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