lecture16 - 16 Lecture 16 16.1 Refraction Let’s have a a...

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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 wave-length 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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