4.5 Images Formed by the Refraction of Light

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1 Figure 89: Practical structure of an optical fibre. Absorption in the glass tube leads to a gradual decrease in light intensity. For optical fibres, the glass used for the core has minimum absorption at a wavelength of 1.3 µm, i.e. in the infra red. Hence, this is the laser wavelength used for long-distance signal transmission ( 10 5 m without significant loss). 4.5 Images Formed by the Refraction of Light A well-known effect in optics is that the apparent depth of an object submerged in a transparent liquid, as seen by an observer in air, appears to be different to the actual depth in the liquid. For example, The apparent depth of a stone on the bottom of a shallow pond appears to be less than its actual depth. A fish in an aquarium (figure ) appears to be closer to the glass wall 76 Knight, Figure 23.26(b), page

2 than it really is. Figure 90: Refraction of light rays causes the apparent depth of an object to change. Consider this second example: Let us ignore the presence of the thin glass wall (this produces only a very small refractive effect). The light rays are moving from a liquid (refractive index n 1 ) to air (refractive index n 2 < n 1 ). The light rays are refracted away from the normal at the water/air boundary. Geometrically, our eye perceives a virtual image of the fish to be formed at a closer distance to the boundary than that of the object itself. 148

3 This last point is a general rule: Objects appear to be closer than they really are because of the refraction of light at the medium/medium boundary. We recall from reflection that rays reflected from a plane mirror appear to diverge from a virtual image. Similarly, in refraction, rays from an object point P refract at a boundary between the two media such that the rays appear to diverge from a point P i.e. a virtual image. Figure shows a boundary between two transparent media with refractive indices n 1 and n 2. Point P, the object, is a source of light. Point P, the virtual image, is the point at which the rays appear to diverge from. Let us define the Object distance, s, to be the distance between the object and the boundary. Image distance, s, to be the distance between the image and the boundary. We want to find s. For construction purposes, we set up the optical axis, a line perpendicular to the boundary that passes through P. Consider a ray that leaves the object at an angle θ 1 with respect to the optical axis. 77 Knight, Figure 23.27, page

4 Figure 91: Finding the position of a virtual image in refraction. 150

5 Suppose that the ray meets the boundary at a point which is a distance l from the optical axis. The light ray thus refracts through the boundary into the second medium such that the angle of refraction is θ 2. Geometrically, we can trace the refracted ray back to the point P, where the angle between this imaginary ray and the optical axis is θ 2. By simple trigonometry, we see that i.e. tan θ 1 = l s, tan θ 2 = l s, (4.9) or l = s tan θ 1 = s tan θ 2, (4.10) But we also recall from Snell s law that s = s tan θ 1 tan θ 2. (4.11) n 2 = n 1 sin θ 1 sin θ 2. (4.12) In reality, when we observe the object, the optical axis is a line joining the object to our eye. In this case, The distance l is the radius of your pupil, i.e. we have that l s, l s. (4.13) The angles θ 1 and θ 2 are necessarily extremely small (the angles in figure 91 have been grossly exaggerated for clarity). 151

6 Mathematically, we can employ the small-angle approximation where θ is in radians. Thus in (4.12), sin θ θ, (4.14) sin θ 1 sin θ 2 = n 2 n 1 θ 1 θ 2 n 2 n 1. (4.15) In a similar manner, we find that the tangent of an angle obeys the small-angle approximation: i.e. if θ is very small and in radians, tan θ θ. (4.16) Therefore, in (4.11), tan θ 1 = s tan θ 2 s θ 1 θ 2 s s. (4.17) Equating (4.15) and (4.17), we then find an expression for the image distance s, namely s = n 2 n 1 s. (4.18) We note from (4.18) that s is independent of θ 1, so that all rays parallel to the axis (so-called paraxial rays) appear to diverge from the same point P. Example: 78 a biologist keeps a specimen of his favourite beetle embedded in a cube of polystyrene plastic. The hapless bug appears to be 3.0 cm within the plastic. What is the beetle s actual distance beneath the surface? 78 Knight, Example 19, page

7 Solution: so parallel rays (i.e. paraxial rays) from the beetle refract into the air and then enter into the observer s eye. The rays in the air when traced back into the plastic appear to be coming from the beetle at a shallower location, a distance s from the plastic-air boundary. If we have an actual object distance s, and an image distance s = 3.0 cm, then using (4.18), i.e. so if s = 3.0 cm, then s = n 2 n 1 s = n air n plastic s, 3.0 = 1.0 s s = 4.77 cm Colour and Dispersion The phenomenon of colour is associated with the wavelength of light our eyes are sensitive to the wavelengths of visible light, and our brains interpret visible light at these particular wavelengths as colour. For example, our brains interpret colours with the following wavelengths: Deepest red: 700 nm. Red: 650 nm. Green: 550 nm. Blue: 450 nm. Deepest violet: 400 nm. Most of the results in optics do not depend upon colour: rays of light corresponding to wavelengths in the red end of the visible spectrum ( red light ) generally pass through materials in the same way as rays of blue light. 153

8 4.6.1 Light Passing Through a Prism Newton: I procured me a triangular glass prism to try there with the celebrated phenomena of colours. Consider figure 92, 79 prism. showing the path of white light refracting through a Figure 92: Refraction of white light through a glass prism. Newton proposed the idea that what we perceive as white light is in fact a mixture of all colours. 79 Pink Floyd, The Dark Side of the Moon 154

9 White light can be separated out into its component colours using a prism (figure ). Similarly, a prism can be used to mix all of these colours back into white light. Figure 93: Dispersion and subsequent recombination of white light through a prism. So how does this effect occur? The refractive index of a material is slightly different for different wavelengths of light i.e. for different colours this effect is known as dispersion. For example, glass has a slightly different refractive index for violet light than for green light or red light. Therefore, since the refractive indices are different, by Snell s law we see that different colours refract through different angles and consequently follow slightly different paths. 80 Knight, Figure 23.28(a), page

10 Figure shows a dispersion curve i.e. a curve showing how the refractive index of a material varies with wavelength for glass. Figure 94: Dispersion curves of two types of glass. Notice, for example, n is larger when λ is shorter. n is smaller when λ is larger. Thus, violet light refracts more than red light Rainbow Figures 95 and gives a basic description of how we observe a rainbow. When the sun is behind you, and a ray of sunlight is incident upon a water droplet in the atmosphere: 81 Knight, Figure 23.29, page Knight, Figure 23.30(a), page

11 Figure 95: Production of a rainbow. Figure 96: Production of a rainbow. Refraction occurs at the first boundary the white light is dispersed into its component colours. Refraction/reflection occurs at the second boundary while most of the rays are refracted out of the droplet and into the air, a 157

12 small number are reflected back into the droplet. Refraction occurs at the third boundary the coloured rays are dispersed even more. So white light enters the droplet and is dispersed; while most of the light is refracted out of the droplet at the second boundary, a small percentage of light ends up being refracted out of the droplet on the same side from which it entered. Figure 95 suggests that violet light should be at the top of the rainbow, while red light is at the bottom. However, in reality, it is the opposite: This is because as shown in figure 96, red light is refracted at 42.5, while violet light is refracted at Consequently, both red and violet light dispersed from the same droplet do not enter your eye. It is a combination of refracted light from different droplets that leads to a rainbow: Since red light is refracted through a greater angle than violet light, you have to look higher in the sky to see the red light - hence red is at the top of the rainbow Coloured Filters and Coloured Objects It is well known that coloured filters, or objects (e.g. coloured glass) appear to refract only light of the same colour why is this? A filter/object will absorb all light falling upon it, and re-radiate only the light corresponding to the particular colour. 158

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