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Unformatted text preview: 17. Lecture 17
17.1 Concave mirror
Continuing with our study of the concave mirror we can ﬁnd now where the image
is formed. In the approximation of small angles we are using, all the rays of light
emanating from the object that are reﬂected in the mirror intersect at a point which is
the position of the image. To ﬁnd such position we need to us at least two rays. This is
illustrated in ﬁg.98. One is the ray that is parallel to the horizontal axis that is reﬂected
toward the focal point. Another one is the one that goes through the focal point which
is reﬂected parallel to the horizontal axis. This is so because the path of the light can
always be backtracked, so if it arrives parallel then goes to the focus implies that if it
comes from the focus it reﬂects parallel. Where these two rays intersect is where the
image is located. Another ray that can be used is the one that hits the center of the
mirror which reﬂect with the same angle as it arrives. This is because at that point
the normal to the mirror is horizontal so a ray hitting the center behaves as if it were
reﬂecting from a vertical ﬂat mirror. Just to be clear when we say “horizontal” or
“vertical” we are assuming that the axis of the mirror is the horizontal direction in the
picture. From the ﬁgure we see that, in this case, the image is real, namely it is formed
in front of the mirror. On the other hand, looking at ﬁgure 99 we ﬁnd that, if the object
is closer to the mirror than the focal distance f , then the image is virtual. In the next
section we will be more precise and ﬁnd a mathematical equation that determines the
position and height of the image. Figure 98: Image of an object by a concave mirror when the object is further than the focal
point. Follow the rays and see how the image is constructed. The highlighted triangles are
similar to each other and used in deriving the mirror equation. – 112 – Figure 99: Image of an object by a concave mirror when the object is closer than the focal
point. 17.2 Mirror equation
Consider the case of the concave mirror that we have already discussed and illustrated
in ﬁg.98. If we put an object of height h at a distance s from the mirror we are
interested in ﬁnding the height h and position s of the image. Before starting let us
discuss how we set up the sign conventions. We are going to take s > 0 if the image
is formed on the same side as the object (real image) and s < 0 if it is formed behind
the mirror (virtual image). Furthermore, if the image is upright we take h > 0 and if
it is inverted we take h < 0.
We can now proceed to ﬁnd the position and height of the image. From ﬁg.98,
using that the highlighted triangles are similar (namely they have the same angles and
their sides are proportional) we ﬁnd that:
h
−h
=
s−f
f (17.1) Here we took into account that h < 0 so the length of the corresponding side of the
triangle is actually −h . If we do the same with the image we ﬁnd another equality:
− h
h
=
−f
s
f – 113 – (17.2) which actually is the same as the previous equation after interchanging s ↔ s , h ↔ −h .
From eq.(17.1) we ﬁnd
hf
h = −
(17.3)
s−f
replacing this value of h in the eq.(17.2) we ﬁnd:
hf
h
=
(s − f )(s − f )
f ⇒ hf
h ( s − f )
=
s−f
f ⇒ s = f2
+f
s−f (17.4) Taking common denominator in the expression for s we ﬁnally ﬁnd
sf
s−f
hf
h = −
s−f
s = (17.5)
(17.6) where we included eq.(17.3). These two equations completely determine the position
and size of the image. Although we derived them for the case where s > f , it is straightforward to check they also work when s < f . It is conventional to rewrite them in a
way that makes more evident the symmetry between the object and the image:
s = sf
s−f ⇒ 1
s−f
11
=
=−
s
sf
f
s ⇒ 1
1
1
+ =
ss
f (17.7) For the height we have: h
hf s − f
h
=−
=−
s
s − f sf
s
So we get the equivalent equations:
1
1
1
+ =
ss
f
−h
h
=
s
s (17.8) (17.9)
(17.10) 17.3 Convex mirror
A convex mirror is shaped as the exterior of a sphere. Rays which arrive parallel to the
axis are divergent after being reﬂected. They appear to originate from a point behind
the mirror called the focal point. On the other hand, a ray that is directed toward the
focal point will be reﬂected parallel. In ﬁg.100 we see that in this case the image is
always virtual and upright. With some algebra and using the same ideas as before we
can derive that the convex mirror obeys the same mirror equation as the concave one
with the only change being that now f < 0. – 114 – Figure 100: Image of an object by a convex mirror. The image is always virtual and upright.
In this case f < 0. 17.4 Convergent lens
Another important optical component is the lens. In the ideal situation, rays arriving
parallel to the axis of the lens are transmitted and converge into a single point called
the focal point. The lens works using the laws of refraction. Typically it is made of
glass or a similar material. The rays of light are refracted twice, once in each surface
of the lens and that produces the desired eﬀect. As with the mirror, the behavior of
the lens is only approximate and valid for rays which are not too far from the axis
and/or arrive at small angles. If the rays arrive from the other side of the lens they will
converge into another focal point situated at the same distance as the one we already
discussed but on the other side of the lens. Furthermore, if a ray goes through a focal
point, it will become parallel to the axis after crossing the lens. Finally, a ray going
through the center of the lens is transmitted right through. The rules to construct
images are similar as with the mirrors. In the ﬁgures 101 and 102 we see two cases,
one where the object is far from the lens and the other when it is closer that f . – 115 – Figure 101: Image of an object by a convergent lens when the object is further than the
focal point. Figure 102: Image of an object by a convergent lens when the object is closer than the focal
point. – 116 – Figure 103: Demo illustrating the properties of lenses – 117 – ...
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