Unformatted text preview: 16. Lecture 16
16.1 Refraction
Let’s have a a look at Snell’s law
n1 sin θ1 = n2 sin θ2 (16.1) from the point of view of waves. First we can deﬁne a wave front as the surface where
the waves have maximum amplitude. Such wave front moves with the speed of light.
A wavelength behind it there is another wave front where the amplitude is maximum
etc., there are in fact many wave fronts. When the wave reaches the interface (see ﬁg.93,
each part of the front does so at a diﬀerent time. Therefore, two arbitrary points A,B
separated by a distance L on the interface are not in phase, namely if A is a maximum
B is not. In fact B is behind A by a time ∆t determined by
c1 ∆t = L sin θ1 (16.2) To see that we can use some geometry to prove that the angle of incidence θ1 is the same
as the other angle labeled as θ1 in the ﬁgure. On the other hand, the same oscillations
of the ﬁeld are seen from the other side of the interface. This implies that the phase
diﬀerence between A and B given by ∆t has to also be equal to
c2 ∆t = L sin θ2
From here we derive that (16.3) L
L
sin θ1 = sin θ2
c1
c2 (16.4) and, canceling L and multiplying by c the speed of light in the vacuum we derive that
c
c
sin θ1 = sin θ2
c1
c2 ⇒ n1 sin θ1 = n2 sin θ2 (16.5) which is Snell’s law. Notice that part of the light is always reﬂected form the interface.
In the case where n1 > n2 (for example light going from water to air) we see that
sin θ2 =
Since n2
n1 n1
sin θ1
n2 (16.6) < 1 there is a critical angle θ1 = θc such that
sin θc = n2
n1 – 106 – (16.7) In that case we get θ2 = pi namely the transmitted light emerges parallel to the
2
interface. For angles θ1 > θc there is no value of θ2 that satisﬁes Snell’s law because
sin θ2 ≤ 1 (sine is always smaller than one). In that case we have a phenomenon known
as total reﬂection, all the light is reﬂected and nothing is transmitted. You can verify
that when you go to a swimming pool and look up while being under water. There is
am angle beyond which you cannot see outside. Figure 93: Refraction of light in the interface between two media. Snell’s law can be derived
from the wave nature of light. The dotted lines represent wave fronts. 16.2 Mirrors
A mirror can be thought as an interface that reﬂects all the incoming light, that is,
nothing is transmitted. There are several types of mirror characterized by their shapes.
16.2.1 Flat mirror
Flat mirrors are the most common ones. As we already discussed, the reﬂected ray
forms the same angle with the normal as the incident ray. This follows form the same – 107 – principles used to derive refraction. if we look at the front wave now the speed of light
is the same so the angle is the same. A consequence of this is that, as we all know, a
ﬂat mirror gives an image identical to the original only that is reversed front to back.
The image appears to be at the same distance behind the mirror as the original object
in front. To see why this is so we can look at ﬁgure 94. If you look at the reﬂected
rays originating from a point A, they appear to come from point A’. So you interpret
this information as if there were another object in point A’ although of course there is
none. This method to analyze the formation of images is applied to more complicated
cases as we discussed now. Figure 94: Reﬂection of light by a ﬂat mirror. The reﬂected rays appear to emanate from
the point A which is the image of point A. 16.2.2 Concave mirror
A concave mirror has the shape of a section of a spherical surface. They are characterized by R the radius of the corresponding sphere (see ﬁg. 95). The center of the
sphere is called the center of curvature of the mirror and R the radius of curvature. If
we look at what happens for a ray that originates in point A and moves parallel to the
horizontal axis in the same ﬁgure, we see that, after being reﬂected it intersects the
horizontal axis at a point which approximately R away from the mirror. Such point is
2
called the focal point. The proof of that is that, from the ﬁgure we have α = 2θ and
then
sin θ = h
R – 108 – (16.8) sin 2θ = h
(16.9) Now we have to use an important formula valid when an angle is small (in radians):
sin θ θ when θ 1 Using this we ﬁnd
θ
2θ h
,
R
h
⇒ = R
2 (16.10) (16.11) From the triangle drawn in the ﬁgure we ﬁnd that the focal point is at the same distance
from the center of curvature. Therefore the focal point is at a distance f R from
2
the mirror. In the approximation of small angles all ray parallel to the horizontal axis
are reﬂected toward the focal point. Figure 95: Reﬂection of light by a concave mirror. A ray parallel to the horizontal axis
is reﬂected toward the focal point at a distance f = R of the mirror. The two segments
2
highlighted in green have the same length (denoted as in the text). – 109 – Figure 96: Demo: Reﬂection of light by a concave mirror. – 110 – Figure 97: Demo: an interesting eﬀect of real images. The image of a small object can be
easily confused with the real thing. – 111 – ...
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 Fall '11
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 Light, Sin, focal point, Geometrical optics

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