Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Jac Phys 1 2E - 02 Page 23 Tuesday, October 21, 2003 1:43 PM Chapter Refracting light 2 Remember Before beginning this chapter, you should be able to: • associate changes in direction of light rays with refraction • describe some everyday uses of lenses. Key ideas After completing this chapter, you should be able to: • describe the bending of light as it passes from one medium into another • use the ray model to describe the refraction of light • mathematically model refraction using Snell’s Law • use the ray model of light to describe and explain total internal reflection and mirages • describe the path of light through an optical fibre • discuss the limitations of the ray model in describing and explaining refraction • demonstrate how a particle model and a wave model can be used to describe and explain the reflection and refraction of light • use the ray model to locate and describe images formed by convex and concave lenses. Figure 2.1 T he path of a light ray bends when travelling from one material into another, such as through water. A person’s legs appear shorter. Inve Jac Phys 1 2E - 02 Page 24 Tuesday, October 21, 2003 1:43 PM ations stig Investigation 2.1 Seeing is believing Refraction is the bending of light as it passes from one medium into another. Experience shows that when you are spearing for fish in the shallows you must aim the spear below where the fish appears to be in the water. At the beach or in a pool, people standing in the shallows appear to have shorter legs. Our perception is distorted, but the reason is not apparent. When we set up a special situation, such as in figure 2.2, where a straight rod is placed in a beaker of liquids that do not mix, the idea of change of direction is apparent. This change in direction of the light is called refraction. Figure 2.2 An example of refraction BENDING OF LIGHT The ray model can help explain our observations of light. If a fish seems closer to the surface of the water, the ray of light from the fish must have bent. To our eye, the ray seems to be coming from another direction (see figure 2.3). Given that light can travel both ways along a light path, the fish will see the spear thrower further towards the vertical than he actually is. air water Fish is here. Fish appears to be here. Figure 2.3 The rays from the fish bend when they enter the air. To the eye, the rays appear to come from a point closer to the surface. The normal is a line that is perpendicular to a surface or a boundary between two surfaces. The angle of incidence is the angle between the incident ray and the normal. The angle of refraction is the angle between the refracted ray and the normal. The ray model not only gives us a way of describing our observations of the bending of light, but also of taking measurements. The angle that a ray of light makes with the normal, angle of incidence and angle of refraction can be measured and investigated. Snell’s Law In 1621, the Dutch physicist Willebrand Snell investigated the refraction of light and found that the ratio of the sines of the angles of incidence and refraction was constant for all angles of incidence. The diagram on page 25 shows how an incident ray is affected when it meets the boundary between air and water. The normal is a line at right angles to the boundary, and all angles are measured from the normal. Some of the light from the incident ray is reflected back into air. The rest is transmitted into the water. The following ratio is a constant for all angles for light travelling from air to water: sin θ --------------------i- = constant. sin θ r 24 WAVE-LIKE PROPERTIES OF LIGHT Jac Phys 1 2E - 02 Page 25 Tuesday, October 21, 2003 1:43 PM angle of incidence normal angle of reflection incident ray boundary reflected ray i i air water r angle of refraction refracted ray sin θ The ratio -----------------------i is constant for all angles for light sin θ r travelling from air to water. Figure 2.4 Relative refractive index is a measure of how much light bends when it travels from any one substance into any other substance. The absolute refractive index of a substance is the relative refractive index for light travelling from a vacuum into the substance. It is commonly referred to as the refractive index. Snell repeated his experiments with different substances and found that the ratio was still constant, but it had a different value. This suggested that different substances bend light by different amounts. (Remember that some light is always reflected.) In fact, there is a different ratio for each pair of substances (for example air and glass, air and water). A different ratio is obtained for light travelling from water into glass. The value of the ratio is called the relative refractive index because it depends on the properties of two different substances. The bending of light always involves light travelling from one substance to another. It is not possible to find the effect of a particular substance on the deflection of light without adopting one substance as a reference standard. Once you have a standard, every substance can be compared with it. A natural standard is a vacuum — the absence of any substance. The absolute refractive index of a vacuum is given the value of one. From this, the absolute refractive index of all other substances can be determined. Some examples are given in table 2.1. (The word ‘absolute’ is commonly omitted and the term ‘refractive index’ usually refers to the absolute refractive index.) Table 2.1 Values for absolute refractive index MATERIAL VALUE Vacuum 1.000 0 Air at 20°C and normal atmospheric pressure 1.000 28 Water 1.33 Perspex 1.49 Quartz 1.46 Crown glass 1.52 Flint glass 1.65 Carbon disulfide 1.63 Diamond 2.42 CHAPTER 2 REFRACTING LIGHT 25 Web Jac Phys 1 2E - 02 Page 26 Tuesday, October 21, 2003 1:43 PM s link Refraction applet The refractive index is given the symbol n because it is a pure number without any units. This enables a more useful restatement of Snell’s Law, for example: nair sin θair = nwater sin θwater. More generally this would be expressed as follows: n1 sin θ1 = n2 sin θ2. This formula is more useful because the two values for a substance — its refractive index and the angle the light ray makes — are together on the same side and are combined in the same way on both sides of the equation. normal Figure 2.5 A graphical depiction of Snell’s Law for any two substances. Note that the light ray has no arrow, because the relationship is true for the ray travelling in either direction. SAMPLE PROBLEM 2.1 Solution: medium 1 refractive index n1 1 boundary medium 2 refractive index n2 2 n 1 sin 1 = n 2 sin 2 A ray of light strikes a glass block of refractive index 1.45 at an angle of incidence of 30°. What is the angle of refraction? nair = 1.0; θair = 30°; nglass = 1.45; θglass = ? 1.0 × sin 30° = 1.45 × sin θglass (substitute values into Snell’s Law) sin 30 ° sin θglass = ----------------(divide both sides by 1.45, the refractive 1.45 index of glass) = 0.3448 (calculate value of expression) (use inverse sine to find angle whose sine ⇒ θglass = 20.17° is 0.3448) (round off to two significant figures) ⇒ θglass = 20°. SAMPLE PROBLEM 2.2 Solution: A ray of light enters a plastic block at an angle of incidence of 40°. The angle of refraction is 30°. What is the refractive index of the plastic? nair = 1.0; θair = 40°; nplastic = ?; θplastic = 30° 1.0 × sin 40° = nplastic × sin 30° (substitute values into Snell’s Law) sin 40 ° nplastic = ----------------sin 30 ° ⇒ = 1.286 nplastic = 1.3. ⇒ (rearrange formula to get the unknown by itself) (calculate value of expression) (round off to two significant figures) AS A MATTER OF FACT L ight can be bent by a strong gravitational field, such as that near the Sun. The gravitational field can act like a convex lens. Light from a distant star that is behind and blocked by the Sun bends around the Sun so that astronomers on Earth see an image of the star to the side of the Sun. 26 WAVE-LIKE PROPERTIES OF LIGHT Jac Phys 1 2E - 02 Page 27 Tuesday, October 21, 2003 1:43 PM Speed of light in glass During the seventeenth century, it was agreed that light changed speed when it travelled from air into glass, but how? One explanation or model of light proposed that light travelled faster in glass. Another model of light proposed that light travelled slower in glass. Scientists at the time did not have the technology to measure the speed of light through materials such as water or glass. It was only in the nineteenth century that Augustin-Jean Fresnel and Jean Bernard Léon Foucault were able to measure the speed of light in water as being less than the speed in air. The second explanation above was accepted and the other rejected. This means that the refractive index now has a physical meaning: speed of light in a vacuum absolute refractive index of water = ----------------------------------------------------------------------------------------------speed of light in water where speed of light in a vacuum = 3.0 × 108 m s−1. SAMPLE PROBLEM 2.3 The refractive index of glass is 1.5. How fast does light travel in glass? 8 3.0 × 10 1.5 = ------------------------------------------------------------(speed of light in glass) Solution: 8 3.0 × 10 ⇒ speed of light in glass = ----------------------1.5 (rearrange formula to get the unknown by itself) = 2.0 × 108 m s−1. Apparent depth At the beginning of the chapter, the experience of the spear thrower was described and explained. A similar phenomenon occurs when a spear thrower is directly above a fish — the fish appears to be closer to the surface than it actually is. This observation is known as apparent depth. Swimming pools provide another example of apparent depth — they look shallower than they actually are. The refraction of light combined with our two-eyed vision makes the pool appear shallower. The relationship is illustrated in figure 2.6 and can be expressed as follows: real depth -------------------------------------------------------- = refractive index. apparent depth air water apparent depth real depth Figure 2.6 The phenomenon of apparent depth CHAPTER 2 REFRACTING LIGHT 27 Jac Phys 1 2E - 02 Page 28 Tuesday, October 21, 2003 1:43 PM 2.4 A swimming pool is 2.00 m deep. How deep does it appear as you stand on the edge of the pool? Solution: real depth = 2.00 m; apparent depth = ?; refractive index of water = 1.33 Inve SAMPLE PROBLEM ations stig Investigation 2.2 Using apparent depth to determine the refractive index 2.00 ---------------------------------------- = 1.33 apparent depth 2.00 ⇒ apparent depth = ----------1.33 (substitute values into relationship) (rearrange formula to get the unknown by itself) = 1.50 m. TOTAL INTERNAL REFLECTION AND CRITICAL ANGLE Light can play some strange tricks. Many of these involve refraction away from the normal and the effect on light of a large increase in the angle of incidence. Figure 2.7 There are no mirrors in a fish tank but strange reflections can be seen. It appears that light is being reflected off the side of the fish tank and the water surface. The critical angle is the angle of incidence for which the angle of refraction is 90°. The critical angle exists only when light passes from one substance into a second substance with a lower refractive index. Total internal reflection is the total reflection of light from a boundary between two substances. It occurs when the angle of incidence is greater than the critical angle. 28 It has already been mentioned that some light is reflected off a transparent surface, while the rest is transmitted into the next medium. This applies whether the refracted ray is bent towards or away from the normal. However, a special situation applies when the refracted ray is bent away from the normal. This is illustrated in figure 2.8. As the angle of incidence increases, the angle of refraction also increases. Eventually the refracted ray becomes parallel to the surface and the angle of refraction reaches a maximum value of 90° (see figure 2.8(b)). The corresponding angle of incidence is called the critical angle. If the angle of incidence is increased beyond the critical angle, all the light is reflected back into the water, with the angles being the same. This phenomenon is called total internal reflection (see figure 2.8(c)). WAVE-LIKE PROPERTIES OF LIGHT Jac Phys 1 2E - 02 Page 29 Tuesday, October 21, 2003 1:43 PM (a) Before critical angle (b) At critical angle (c) After critical angle (total internal reflection) r air water c c i i Figure 2.8 Three stages of refraction leading to total internal reflection The critical angle can be calculated using Snell’s Law (see the following sample problem). SAMPLE PROBLEM 2.5 Solution: What is the critical angle for water given that the refractive index of water is 1.3? nair = 1.0; θair = 90°; nwater = 1.3; θwater = ? 1.0 × sin 90° = 1.3 × sin θwater (substitute data into Snell’s Law) sin 90 ° (rearrange formula to get the unknown ⇒ sin θwater = ----------------1.3 by itself) = 0.7692 ⇒ (determine sine values and calculate expression) θwater = 50.28° (use inverse sine to find angle) θwater = 50°. (round off to two significant figures) Total internal reflection is a relatively common atmospheric phenomenon (as in mirages) and it has technological uses (for example, in optical fibres). Mirages There are several types of mirage that can be seen when certain atmospheric conditions enable total internal reflection to occur. These mirages appear because the refractive index of air decreases with temperature. A common type of mirage occurs in the desert or above a road on a sunny day. As displayed in figure 2.10, at ground level the air is hot (A) with a refractive index close to 1 (B). As height increases, the temperature of the air decreases (C) and its refractive index increases (D). Figure 2.9 Mirages such as this are common on hot, sunny days. CHAPTER 2 REFRACTING LIGHT 29 Jac Phys 1 2E - 02 Page 30 Tuesday, October 21, 2003 1:43 PM refractive index temperature high A D temperature increasing 1.000 28 (normal air) refractive index increasing B C low 1.0 (vacuum) ground level Figure 2.10 height tree level Temperature and refractive index profiles for the mirage phenomenon Rays of light from a car, for example, go in all directions. The air above the ground can be considered as layers of air. The closer to the ground, the higher the temperature and the lower the refractive index. As a ray moves into hotter air, it bends away from the normal. After successive deflections, the angle of incidence exceeds the critical angle for air at that temperature and the ray is totally internally reflected. As the ray emerges, it follows a similar path, refracting towards the normal as it enters cooler air. An image of the car can be seen below street level (see figures 2.9 and 2.11). The mirage is upside down because light from the car has been totally internally reflected by the hot air close to the road surface. Figure 2.11 Web The mirage of the car appears upside down due to total internal reflection in the hot air close to the ground. s link Mirages and more warm air hot air road Another mirage that depends on layers of air at different temperatures is known as the ‘Fata Morgana’ in which vertical streaks, like towers or walls, appear. This occurs where there is a temperature inversion — very cold at ground level and warmer above — and very stable weather conditions. The phenomenon is named after Morgan le Fay (Fata Morgana in Italian) who was a fairy and half-sister to King Arthur of the Celtic legend. She used mirages to show her powers and, in the Italian version of the legend, lived in a crystal palace under the sea. The mirage is often seen in the Strait of Messina and over Arctic ice. As shown in figure 2.12, the light rays from a distant point are each refracted by the different layers of air, arriving at different angles to the eye. The effect is that the point source (P) becomes a vertically extended source, like a tower or wall (see figure 2.13). rough sea P ice Figure 2.12 Ray paths for the Fata Morgana 30 WAVE-LIKE PROPERTIES OF LIGHT Jac Phys 1 2E - 02 Page 31 Tuesday, October 21, 2003 1:43 PM Figure 2.13 An example of the Fata Morgana. The conditions that encourage the Fata Morgana are particularly common in the polar regions over ice. Optical fibres Web An optical fibre is a thin tube of transparent material that allows light to pass through without being refracted into the air or another external medium. Another example of total internal reflection is in the important technological application of optical fibres. Optical fibres have become a feature of modern life. A thin, flexible cable containing an optical fibre can be placed inside a person’s body to transmit pictures of the condition of organs and arteries, without the need for invasive surgery. The same can be done in industry when there is a problem with complex machinery. Optical fibres are also the basis of the important telecommunications industry. They allow high quality transmission of many channels of information in a small cable over very long distances and with negligible signal loss (see figure 2.14). s link Fibre optics Figure 2.14 A bundle of optical fibres. Each fibre in the bundle carries its signal along its length. If the individual fibres remain in the same arrangement, the bundle will emit an image of the original object. An optical fibre is like a pipe with a light being shone in one end and coming out of the other. An optical fibre is made of glass which is about 10 micrometres (10 × 10−6 m) thick. Light travels along it as glass is transparent, but the fibre needs to be able to turn and bend around corners. The optical fibre is designed so that any ray meeting the outer surface of the glass fibre is totally internally reflected back into the glass. As shown in figure 2.15, the light ray meets the edge of the fibre at an CHAPTER 2 REFRACTING LIGHT 31 Jac Phys 1 2E - 02 Page 32 Tuesday, October 21, 2003 1:43 PM angle of incidence greater than the critical angle and is reflected back into the fibre. In this way, nearly all of the light that enters the fibre emerges at the other end. optical fibre light ray Figure 2.15 A light ray travels along an optical fibre through total internal reflection. If the glass fibre is exposed to the air, the critical angle for light travelling from glass to air is 42°, which is quite small. Any angle of incidence greater than this angle will produce total internal reflection. If the fibre is very narrow, this angle is easily achieved. However, in both medical and telecommunication uses, fibres are joined in bundles with edges touching. The touching would enable light rays to pass from fibre to fibre, confusing the signal. To overcome this, a plastic coating is put around the glass to separate the glass fibres. The total internal reflection occurs between the glass and the plastic. The critical angle for light travelling from glass to plastic is 82°. This value presents a problem because light meeting the edge of the glass at any angle less than 82° will pass out of the fibre (see figure 2.16). This has implications for the design of the optical fibre and the beam of light that enters the fibre. The fibre needs to be very narrow and the light entering the fibre has to be a thin beam with all the rays parallel. light rays 82° optical fibre Figure 2.16 Light rays entering the fibre at too sharp an angle are refracted out of the fibre. LIMITATIONS OF THE RAY MODEL So far in this chapter we have used the ray model to describe how light is refracted. Ray diagrams illustrate Snell’s Law and have allowed us to visualise a range of optical phenomena such as mirages and to develop technologies such as optical fibres. However, the ray model, which views light as a pencil thin beam, does not offer an explanation of why light refracts. More sophisticated models are needed to provide an explanation for refraction, and in doing so they suggest further experiments to investigate the properties of light more deeply, and to develop new technologies. 32 WAVE-LIKE PROPERTIES OF LIGHT Jac Phys 1 2E - 02 Page 33 Tuesday, October 21, 2003 1:43 PM Two very different models of light were developed in the seventeenth century one by Sir Isaac Newton (1642–1727) in England and the other by Christiaan Huygens (1627–1695) in Holland. Newton’s model was described as a ‘particle model’. In his model, light consists of a stream of tiny, mass-less particles he called corpuscles. The particles stream from a light source like water from a sprinkler. Huygens proposed a wave model of light, where light travels in a similar way to sound and water waves. Light leaves a source in the same way that water ripples move out from a dropped stone. The disturbance of the water surface travels outwards from the source. Figure 2.17 Huygens proposed that light travelled outwards from a source like circular ripples on a pond. HOW DO THE TWO MODELS EXPLAIN THE PROPERTIES OF LIGHT? How light travels Figure Newton’s particle model Once ejected from a 2.18 Reflection of light Every point in the wavefront is a Source source of a small S wavelet. The new wavefront is the envelope of all the wavelets. Web light source the particles continue in a straight line until they hit a surface. Huygens’s wave model Huygens proposed a basic principle: ‘Every point in the wavefront is a source of a small wavelet. The new wavefront is the envelope of all the wavelets.’ s link Huygens’s principle applet Newton’s particle model As particles approach a surface they are repelled by a force at the surface that slows down and reverses the normal component of the particle’s velocity, but does not change its tangential component. The particle is then reflected from the surface at an angle equal to its angle of approach. The same process happens when a billiard ball hits the cushion. Figure 2.19 Newton’s particle model of reflection i i´ mirror Huygens’s wave model As each part of the wavefront arrives at the surface, it produces a reflected wavelet. The new wavelets overlap to produce the next wavefront, which is travelling away from the surface at an angle equal to its angle of approach. D B Figure 2.20 The wave model of reflection. C and D are parallel, incoming rays. AB is the wavefront. When A hits the mirror a circular wavelet is produced. By the time B has reached the mirror at E, the reflected wavelet has travelled out to F. The line EF is the reflected wavefront. E C F mirror A (continued ) CHAPTER 2 REFRACTING LIGHT 33 Jac Phys 1 2E - 02 Page 34 Tuesday, October 21, 2003 1:43 PM Refraction of light Newton’s particle model In approaching a denser medium, the particles experience an attractive force which increases the normal component of the particle’s velocity, but does not affect the tangential component. This has the effect of changing the direction of the particles, bending them towards the normal where they are now travelling faster in the denser medium. Snell’s Law can be explained by this model. Inve i Figure 2.21 The particle model of refraction. The particles are pulled towards the denser medium, resulting in a change in direction. ations stig air water r Investigation 2.3 Refraction of particles Huygens’s wave model When the wavefront meets a heavier medium the wavelets do not travel as fast as before. This causes the wavefront to change direction. In this case the wavefront bends towards the normal when it enters a medium where the wave is slowed down. Snell’s Law can be explained by this model. B C E A F The wave model of refraction. C and D are parallel, incoming rays. AB is the wavefront. When A hits the surface a circular wavelet of slower speed and so smaller radius is produced. By air the time B has reached the surface at E, the refracted water wavelet has only gone as far as F. The line EF is the refracted wavefront, heading in a direction bent towards the normal compared to the incoming wavefront, AB. A point of difference Now, with these two explanations of refraction, there is a clear distinction between the two models. When light bends towards the normal as it enters water (a denser medium), the particle model says it is because light travels faster in water (the denser medium), whereas the wave model says it is because the light is travelling slower. In the seventeenth century they did not have the technology to measure the speed of light in water. However, the particle model became the accepted explanation, partly because of Newton’s status, and partly because Huygens’s principle suggested that light should bend around corners like sound, and there was no evidence of this at the time. (Newton himself actually thought that the particles in his model needed to have some wavelike characteristics to explain some of his other observations of light and colour.) New evidence emerges In 1802, Thomas Young (1773–1829) showed that in fact light could bend around an edge. This led to the rejection of the particle model, as it had no mechanism to explain how particles could bend around a corner, and consequently the wave model re-emerged. In 1856, Foucault measured the speed of light in water and showed that it was indeed slower than that in air, clinching the argument for the wave model. 34 WAVE-LIKE PROPERTIES OF LIGHT Inve Figure 2.22 D ations stig Investigation 2.4 Refraction of waves Jac Phys 1 2E - 02 Page 35 Tuesday, October 21, 2003 1:43 PM FORMING IMAGES WITH LENSES Inve A convex lens is a lens that is thicker in the middle than at the edges. A concave lens is a lens that is thinner in the middle than at the edges. Converging lenses are those that bend incident parallel rays towards a focus. That is, they converge light. Diverging lenses are those that bend incident parallel rays away from each other. That is, they diverge light. ations stig The refraction of light at a boundary can be put to use if the boundaries are curved. There are two possibilities for curved boundaries — curving inwards or curving outwards. A convex lens has its faces curving outwards. A lens that curves inwards is a concave lens. The simple ray tracing in figure 2.23 illustrates what each lens does to the light rays. As rays enter the glass they are bent towards the normal. When they reach the air on the other side of the lens, they are bent away from the normal. In the case of the convex lens, the emerging rays come together or ‘converge’ at a point called the focus (F). For the concave lens, the rays move apart or ‘diverge’, so that they appear to come from a point, also called the focus, on the other side of the lens. For these reasons, convex lenses are sometimes called converging lenses, and concave lenses can be called diverging lenses. (a) Convex lens Investigation 2.5 The convex lens as a magnifying glass (b) Concave lens F F Figure 2.23 Refraction of rays through (a) a convex and (b) a concave lens focal plane F Figure 2.24 principal axis Rays converge at a point on the focal plane, which passes through the focus point (F). In fact, the focus is more than just a point. It is a plane — a focal plane through the focus point. In figure 2.24, for example, parallel light rays from a distant object coming in at an angle to the lens are still brought to a focus, not at the focus but elsewhere in the focal plane. This is of use in designing telescopes. Convex lenses — locating images using ray tracing There are similarities in the nature of images produced by a convex lens and those produced by a concave mirror. The features of a convex lens are illustrated in figure 2.25. The convex lens has two symmetrical curved surfaces, which means that it has two focus points. principal axis convex lens pole focus Figure 2.25 focus The convex lens has two focus points. CHAPTER 2 REFRACTING LIGHT 35 Web Jac Phys 1 2E - 02 Page 36 Tuesday, October 21, 2003 1:43 PM The ray model can now be used to locate an image (see figures 2.26 and 2.27). s link Convex lens applet object 3 convex lens image 1 Inve 4 F ations stig 2 Investigation 2.6 Describing images produced by a convex lens convex lens 2 object image F F 1 3 Figure 2.27 The object is inside the focus. This makes the rays diverge after passing through the lens. ‘Backtracking’ these rays reveals that they appear to be coming from a point behind the object. The image is located at this point. eMo F ling del Figure 2.26 The location of the image is determined according to the point where the three rays cross. All the rays that pass through the lens pass through the image. The ray diagram in figure 2.26 demonstrates the following: • Ray 1 leaves the head of the object parallel to the principal axis, is refracted by the lens and passes through the focus on the other side of the lens. • Ray 2 passes through the focus on the same side as the object, is refracted by the lens and emerges travelling parallel to the principal axis. • Ray 3 travels towards the centre of the lens. If the angle of the incoming ray is small and the lens is considered thin, then the ray would appear to continue on in the same direction. All three rays pass through the same point. This is where the image of the head of the object is located. Note that the image could have been located with any two of the rays, but the other can be used to confirm the original location of the image. The features of the image can now be described — location, size, orientation and nature (see chapter 1, page 12). Convex lenses are used in a variety of applications. There are different applications for each location of the object. A range of these applications is given in table 2.2 (see page 38). Model of a convex lens Convex lenses — using a formula to locate and describe images u Ho image object F F f Hi v Figure 2.28 Representation of the elements of a formula to locate images in a convex lens 36 The similarities between the ray diagrams for the concave mirror and the convex lens suggest that the same formula (see below) can be used for both, but with a new sign convention which reads: Real image distances are expressed as positive quantities and virtual image distances are expressed as negative quantities. 1 1 1 --- + --- = --u v f height of image ( H i ) v magnification (M) = ------------------------------------------------------- = --height of image ( H o ) u WAVE-LIKE PROPERTIES OF LIGHT Jac Phys 1 2E - 02 Page 37 Tuesday, October 21, 2003 1:43 PM SAMPLE PROBLEM 2.6 Solution: Skill cks che Significant figures (p. 493) Skill cks che Describe fully the image of a 4.0 cm object 15 cm in front of a convex lens with a focal length of 10 cm. u = 15 cm; v = ?; f = 10 cm; M = ?; Ho = 4.0 cm; Hi = ? 1 1 1 ------ + --- = -----(substitute data into the lens formula) 15 v 10 1 1 1 --- = ------ − -----⇒ (rearrange to get the unknown by itself) v 10 15 1 3 2 = ------ − ------ = ------ (combine fractions with lowest common 30 30 30 denominator) ⇒ v = 30 (invert fractions to get v as numerator) The image is 30 cm from the lens and on the opposite side of the lens to the object. Hi Magnification = ------Ho v = --u Using spreadsheets (p. 504) H 30 -------i = -----(substitute values into magnification formula) 4.0 15 H -------i = 2.0 4.0 ⇒ Hi = 2.0 × 4.0 (rearrange to get the unknown by itself) = 8.0 (calculate expression) The image is 8.0 cm high, real, inverted and located 30 cm from the lens. eMo ⇒ ling del Inve Using a spreadsheet to locate the image from a slide projector ations stig Web Investigation 2.7 Locating an image for a concave lens s link Concave lenses — locating images using ray tracing Concave lenses curve inwards so they are thinner in the middle, as opposed to the convex lens which is thicker in the middle. This means that the rays of light are bent away from the principal axis at both the front and the back surfaces of the lens (see figure 2.23(b), page 35. For this reason they are also called diverging lenses. Concave lenses have a principal focus, like object convex lenses, but it is located on the same side of the lens as the object and is called a virtual focus. Light rays parallel to the principal axis spread out after passing through the concave lens as if F F they are coming from the virtual focus. image concave lens Figure 2.29 The rays diverge for all object positions. An upright, diminished, virtual image is located between the focus and the lens. Concave lens applet Concave lenses — using a formula to locate and describe images As with the formula method with the convex mirror, the focal length of a concave lens can be represented as a negative number because it is a virtual focus. CHAPTER 2 REFRACTING LIGHT 37 Jac Phys 1 2E - 02 Page 38 Tuesday, October 21, 2003 1:43 PM SAMPLE PROBLEM 2.7 Solution: Locate and describe the image formed by a 6.0 cm high object 15 cm in front of a concave lens with a focal length of 10 cm. Ho = 6.0 cm; Hi = ?; u = 15 cm; v = ?; f = −10 cm 1 1 1 Using --- + --- = --u v f 1 1 1 ------ + --- = ---------⇒ 15 v −10 1 1 ⇒ = − ------ − -----10 15 (−3 − 2) = ---------------------30 (substitute data into the lens formula) (rearrange to get the unknown by itself) (combine fractions with lowest common denominator) –5 = -----(simplify expression) 30 30 ⇒ v = -----(invert to get v in the numerator) −5 = −6.0. The image is located 6.0 cm in front of the lens. It is virtual and upright. Hi v Magnification = ------- = --Ho u H 6.0 -------i = ------- = 0.4 (substitute data into magnification 6.0 15 formula) Hi = 0.4 × 6.0 = 2.4. (rearrange to get the unknown by itself and calculate expression) The image is 2.4 cm high. Because they spread light rays further apart, concave lenses do not form real images. Therefore, they are usually used in combination with a convex lens. Table 2.2 Simple applications of convex lenses LOCATION OF OBJECT USES DESCRIPTION OF IMAGE Very large distance away from lens Objective lens of refracting telescope Real, inverted, diminished and located near the opposite focus Beyond twice the focal length from lens Human eye; camera Real, inverted, diminished and located on other side between one and two focal lengths from lens At twice the focal length from lens Correction lens for terrestrial telescope Real, inverted, same size and located two focal lengths from lens Between twice the focal length from lens and the focus Slide projector; objective lens of microscope Real, inverted, magnified and located on other side of lens beyond two focal lengths At the focus Searchlight; eyepiece of refracting telescope No image. The emerging parallel rays do not meet. Between focus and lens Magnifying glass; eyepiece lens of microscope; spectacles for long-sightedness Virtual, upright, magnified and located on same side of the lens and further away 38 WAVE-LIKE PROPERTIES OF LIGHT Jac Phys 1 2E - 02 Page 39 Friday, October 24, 2003 2:08 PM PHYSICS IN FOCUS Inve Flat lenses? ations stig Investigation 2.8 Examining a flat lense lens works by changing the direction of the light ray at the front surface and then again at the back surface. The glass in the middle is there to keep the two surfaces apart. Augustin-Jean Fresnel devised a way of making a lens without the need for all the glass in the middle. The glass surface of the lens is a series of concentric rings. Each ring has the slope of the corresponding section of the full lens, but its base is flat. The slopes of the rings get flatter towards the centre. A Figure 2.30 A side view of a convex Fresnel lens showing how it is constructed This design substantially reduces the weight of the lens, so lenses of this type are used in lighthouses. Their relative thinness means they are also used where space is at a premium, such as in overhead projectors, and as a lens to be used with the ground-glass screens in camera viewfinders. Flat lenses, or Fresnel lenses as they are called, are now attached to the rear windows of vans and station wagons to assist the driver when reversing or parking. MICROSCOPES: AN APPLICATION OF LENSES A convex lens can be used as a simple magnifying glass, but to obtain greater magnification another lens is needed. eMo Compound microscope ling del Using a spreadsheet to model a microscope Skill cks che Using spreadsheets (p. 506) In its simplest form, a compound microscope has two convex lenses, one fatter and with a shorter focal length than the other. The fatter one is located near the object to be magnified and is called the objective lens. The other lens is close to the eye and is called the eyepiece (see figure 2.31). An object placed just outside the focus of the objective lens produces a real, inverted, magnified image. The eyepiece lens is adjusted so that this image is formed inside the focus of the eyepiece lens. The real, inverted image now becomes the object for the eyepiece lens. Since it is inside the focus of the eyepiece, it produces a final magnified image which is virtual and inverted compared with the original object. CHAPTER 2 REFRACTING LIGHT 39 Jac Phys 1 2E - 02 Page 40 Tuesday, October 21, 2003 1:43 PM Inve objective lens eyepiece lens convex lenses ations stig object F F •o Investigation 2.9 Calculating with a microscope •o • Fe Fe • I1 I2 Figure 2.31 Ray diagram for a compound microscope HUMAN AND ANIMAL VISION The human eye is an extraordinary device. It is able to respond to an enormous range of light brightness. The strongest light that the eye can safely detect is 10 000 million times as bright as the weakest light it can detect. It is able to focus on objects from many billions of kilometres away to objects a few centimetres away. It can also detect colour. The parts of the human eye are shown in figure 2.32. ciliary muscle iris (coloured) aqueous humour pupil cornea fovea centralis vitreous humour macula optic nerve blindspot lens sclera Cross-section of a human eye choroid retina Figure 2.32 The retina takes the role of the screen of the eye. It is covered with nerve cells that detect the brightness and colour of the light falling on it. The cornea is the curved front surface of the eye. It refracts light towards the pupil so that it can pass through towards the lens. 40 The purpose of the eye is to produce a sharp, real image on a screen — the retina. Light passes through many refracting elements in the human eye on its way to the retina (see table 2.3). It achieves a focus in two stages. The curved cornea at the front of the eye does two-thirds of the focusing, due to the large difference in refractive index between the cornea and the air. The lens does the fine focusing as there is little difference in the refractive index of the lens and that of the liquid on either side of it. In the rest of the eye, the differences in refractive index are marginal, but they are crucial when the eye wishes to produce clear images of far and near objects. WAVE-LIKE PROPERTIES OF LIGHT Jac Phys 1 2E - 02 Page 41 Tuesday, October 21, 2003 1:43 PM Table 2.3 Refractive index of the parts of the eye PART OF THE EYE REFRACTIVE INDEX Tears 1.33 Cornea 1.37 Aqueous humour 1.33 Lens cover 1.38 Lens centre 1.41 Vitreous humour 1.33 AS A MATTER OF FACT T Web he eye is an amazing instrument. It has an automatic aperture adjustment called the iris. The act of blinking operates the cornea’s built-in scratch remover, lens cleaner and lubricator. In dim light, the eye operates as a supersensitive, black and white television camera. It allows us to see objects with less than 0.1 per cent, or onethousandth, of the light we need for colour vision. The optic nerve packages the visual information from the retina so that the brain receives about 30 discrete ‘frames’ per second in a similar way to television and cinema. s link Modelling the eye applet Accommodation is the adjustment to the thickness of the lens in the eye to ensure that the image on the retina is sharp. When the thickness of the lens changes, so does its focal length. The near point of your eye is the closest an object can be to your eye so that your eye can produce a sharp image of the object on its retina. Accommodation mechanisms Consider the light from a light globe that passes through a convex lens onto a screen to produce an image of the globe’s filament. If the lens is moved to a new position, the screen needs to be moved to obtain a sharp image. The human eye can produce a sharp image on its ‘screen’ (the retina) of objects at various distances. But the eye’s screen stays put, so something else has to change to achieve a sharp image. The only thing that can change is the focal length of the lens. Convex lenses form images of very distant objects at their focus. As objects get closer to the lens, the image is formed behind the focus. For the human eye with its fixed ‘screen’, this means that the focal length must shorten to keep a sharp image on the retina. This adjustment is called accommodation. How short can the focal length of the lens in an eye become? When you bring an object closer to your eye from an arm’s length away, there is a point at which the object becomes fuzzy in appearance. This point is called your near point. As you age, your near point becomes further away (see table 2.4). That is why some older people hold the newspaper further away to read it. Table 2.4 Inve AGE (YEARS) ations stig Investigation 2.10 Measuring your near point Average near points by age NEAR POINT (CM) AGE (YEARS) NEAR POINT (CM) 10 7 40 22 15 8.5 50 40 20 10 60 65 30 15 70 200 CHAPTER 2 REFRACTING LIGHT 41 Web Jac Phys 1 2E - 02 Page 42 Tuesday, October 21, 2003 1:43 PM s link Animal eyes To enable the focal length of the eye to change, the lens has to change its shape. To achieve a shorter focal length, the lens needs to become fatter. This raises important questions. How does the lens change shape? Which is the natural rest position of the lens in the eye? Is the lens relaxed in the short focal length position and needs to be ‘stretched’ to see distant objects? Or is it relaxed in the long focal length position and needs to be ‘squashed’ to see near objects? Muscles can act only by contraction, so how are the muscles arranged in the eye to squash the lens? In fact, the relaxed human eye has a long focal length located at the retina to form images of very distant objects. Ligaments are attached to the outside edge of the lens which are continually pulling outwards; however, around the lens itself is a circular muscle. When this circular muscle contracts, it produces a smaller circle, making a lens that is smaller in diameter, but fatter front to back. Do the eyes of animals work in the same way as human eyes? What lens mechanism would be appropriate for an animal in the wild? In fact, most animals’ eyes are relaxed in the long focal length position because, if the animal is roused from sleep by the sound of a distant predator, the eye is ready to focus on it. While many animals use an adjustable focal length lens like humans, some use other strategies. One strategy is to move the lens backwards and forwards. In most fish and snakes, the lens moves within the eyeball in the same way that a camera lens is moved to produce a sharp image on film. Correcting eye defects Sometimes our eyes do not work properly and corrective lenses need to be prescribed. Corrective lenses were the first prosthetic devices to be invented (in about 1300) but, until early in the twentieth century, they were available only to the wealthy. The two main eye defects are hypermetropia and myopia. PHYSICS IN FOCUS The power of the lens O ptometrists and opticians describe the focusing ability of a lens in terms of its power. The power of a simple lens is defined as: 1 Power = --- . f The unit of the power of a lens is the dioptre (D). For example, a concave lens of a focal length of 25 cm has a power given by: 1 1 Power = --- = -------------- = −4.0 D. f – 0.25 Myopia, or short-sightedness, means that the person’s vision of nearby objects is good, but distant objects are unclear. Parallel rays of light from distant objects are brought to a focus in front of rather than on the retina. The eyeball is too large for the ‘relaxed focal length’ of the eye (see figure 2.33). This defect is not usually noticed until the eyeball approaches its final size in adolescence. It can be overcome by placing a concave lens in front of the eye that spreads the rays apart slightly before they enter the eye (see figure 2.34). 42 WAVE-LIKE PROPERTIES OF LIGHT Jac Phys 1 2E - 02 Page 43 Tuesday, October 21, 2003 1:43 PM Figure 2.33 Myopic vision causes light rays to focus in front of the retina. (a) Spectacle lens F Figure 2.34 (b) Contact lens F Corrective concave lenses are used in spectacles and contact lenses. Figure 2.35 Contact lenses can be a solution to eye defects. They alter the curvature of the front of the eye. Sometimes myopia can be caused by excessive curvature of the cornea, the principal focusing agent in the eye. Recent medical developments use low temperature ultraviolet lasers to ‘shave off’ some tissue from the front of the cornea, to flatten the curvature. Hypermetropia, or long-sightedness, is the reverse of myopia. The lens produces an image of near objects behind instead of on the retina (see figure 2.36). Near objects are not clear, while distant objects are in focus. The muscles around the lens cannot contract enough to shorten the focal length to bring the image onto the retina. The eye needs help to achieve this, and an extra convex lens can help (see figure 2.37). Figure 2.36 Hypermetropia causes light rays to focus behind the retina. This produces a ‘fuzzy’ image. CHAPTER 2 REFRACTING LIGHT 43 Jac Phys 1 2E - 02 Page 44 Tuesday, October 21, 2003 1:43 PM (a) Spectacles (b) Contact lens Figure 2.37 Convex lenses are used in spectacles and contact lenses to correct hypermetropia. As people age, the muscles around the lens slowly weaken and the lens becomes thicker and less flexible. The loss in flexibility means that the near point moves further away and the newspaper needs to be held out to be read. The thickening of the lens, even when the muscles are relaxed, means that distant objects start becoming fuzzy. This means that the eye is showing signs of myopia for distant objects and hypermetropia for near objects, which require different corrective solutions. Hence the need for bifocal lenses, which have a convex surface at the bottom to look down to read the newspaper and a concave surface at the top to look up to drive the car (see figure 2.38). lens for watching TV and driving lens for reading Figure 2.38 44 WAVE-LIKE PROPERTIES OF LIGHT Example of a bifocal lens design Jac Phys 1 2E - 02 Page 45 Tuesday, October 21, 2003 1:43 PM • Light bends as it travels from one medium to another. A measure of a medium’s capacity to bend light is given by its refractive index. • If light travels into a medium of a higher refractive index, the light is bent towards the normal. If light travels into a medium of a lower refractive index, the light is bent away from the normal. This change in direction is summarised in Snell’s Law. Snell’s Law can be expressed as n1 sin θ1 = n2 sin θ2. • When light travels into a medium of a lower refractive index, there will be an angle of incidence for which the angle of refraction is 90°. This angle of incidence is called the critical angle. For angles of incidence greater than the critical angle, all the light is reflected back into the medium. This phenomenon is called total internal reflection. • Two improvements on the ray model of light are the particle model and the wave model. Both of these models explain various light phenomena, but they differ in their prediction of the speed of light in a medium. The wave model’s prediction was found to be correct. • When refractive materials are shaped into convex and concave lenses, both real and virtual images can be formed. • The ray model explains the formation and properties of images in convex and concave lenses. • The formation of images in convex and concave lenses can be mathematically modelled 1 1 1 v with the equations --- + --- = --- and M = --- . u v f u • The formation and properties of images in optical devices such as optical fibres, microscopes and spectacles can be modelled mathematically or by ray tracing. QUESTIONS Understanding 1. What is the angle of refraction in water (n = 1.33) for an angle of incidence of 40°? If the angle of incidence is increased by 10°, by how much does the angle of refraction increase? 2. A ray of light enters a plastic block at an angle of incidence of 55° with an angle of refraction of 33°. What is the refractive index of the plastic? 3. A ray of light passes through a rectangular glass block with a refractive index of 1.55. The angle of incidence as the ray enters the block is 65°. Calculate the angle of refraction at the first face of the block, then calculate the angle of refraction as the ray emerges on the other side of the block. Comment on your answers. 4. Immiscible liquids are liquids that do not mix. Immiscible liquids will settle on top of each other, in the order of their density, with the densest liquid at the bottom. Some immiscible liquids are also transparent. (a) Calculate the angles of refraction as a ray passes down through immiscible layers as shown in the figure below. light ray 25° air n = 1.00 acetone n = 1.357 glycerol n = 1.4746 n = 1.4601 carbon tetrachloride glass beaker n = 1.53 (b) If a plane mirror was placed at the bottom of the beaker, calculate the angles of refraction as the ray reflects back to the surface. Comment on your answers. 5. Calculate the critical angle for light travelling through a diamond (n = 2.5) towards the surface. 6. A ray travelling through water (n = 1.33) approaches the surface at an angle of incidence of 55°. What will happen to the ray? Support your answer with calculations. 7. (a) Calculate the refractive index of the glass prism shown in the figure below so that the light ray meets the faces at the critical angle. Is this value of the refractive index the minimum or maximum value for such a reflection? (b) Draw two parallel rays entering the block. How do they emerge? 8. Calculate the refractive index of the plastic coating on an optical fibre if the critical angle for glass to plastic is 82.0° and the refractive index of glass is 1.500. 45° CHAPTER 2 REFRACTING LIGHT 45 CHAPTER REVIEW SUMMARY Jac Phys 1 2E - 02 Page 46 Tuesday, October 21, 2003 1:43 PM 9. When light passes from air into a glass block, some light is reflected off the surface as well. (a) Describe how the wave model of light explains this observation. (b) Why can’t the particle model explain this observation? 10. Use ray tracing to determine the full description of the following objects: (a) a 4.0 cm high object, 20 cm in front of a convex lens with a focal length of 15 cm (b) a 3.0 mm high object, 10 cm in front of a convex lens with a focal length of 12 cm (c) a 5.0 cm high object, 200 cm in front of a convex lens with a focal length of 10 cm. 11. A bull’s eye has a diameter of about 5.0 cm. The bull needs a clear image of the food it is eating, so its near point is about 30 cm. Calculate the minimum focal length of the lens in the bull’s eye. 12. What does ‘accommodation mechanism’ mean? Give an example. Application 13. A ray of light enters a parallel-sided glass block (n = 1.55) at an angle of incidence of 35°, as shown in the figure below. (d) Some or all of the light will be reflected from the adjacent face. Draw the subsequent path of this light, showing angles. (e) Would your answer to (c) be different if the initial angle of incidence was larger or smaller than 35°C? (Hint: Assume the ray meets the adjacent face at the critical angle and then work back to the initial angle of incidence.) (f) Does the answer to (e) depend on the refractive index of the material? Set up the spreadsheet (shown at the base of the page) to investigate how the initial angle of incidence varies with the refractive index if the ray emerges from the adjacent face at 90°. Increase the refractive index in small steps to find the value for which the angle of incidence becomes 90°. What is unusual about this value for the refractive index? 14. Describe what a diver would see when looking up at a still water surface. 15. A right-angled glass prism (n = 1.55) is placed under water (n = 1.33), as shown in the figure below. A ray of light enters the longest side along the normal. What happens to the ray of light? air CHAPTER REVIEW 35° 5 cm water n = 1.55 (a) Calculate the angle of refraction at the top surface and the angles of incidence and refraction as the ray emerges at the bottom surface. (b) If the initial ray (at the same angle) entered the block further to the right, the emerging ray would hit the corner of the block. If the block is 5.0 cm wide, how far from the edge should the initial ray enter the block for this to occur? (c) If the initial ray is moved even closer to the edge, the emerging ray will meet the adjacent side edge. What angle of incidence does the light ray make with the adjacent face? Will the ray pass out of the block or will it be totally internally reflected? 46 WAVE-LIKE PROPERTIES OF LIGHT 45° 16. A fish looking up at the surface of the water sees a circle, inside which it sees the ‘air world’. Outside the circle it sees the reflection of the ‘water world’. If the fish is 40 cm below the surface, calculate the radius of the circle (nwater = 1.33). 17. Light enters an optical fibre 1.0 µ m in diameter, as shown in the figure on page 47. Some light goes straight down the centre. Another ray is angled, leaving the central line and meeting the outside edge at slightly more than the critical angle of 82° then reflects back to the central line. (a) How much further did this ray travel? (b) Calculate the speed of light in the glass and determine the Jac Phys 1 2E - 02 Page 47 Tuesday, October 21, 2003 1:43 PM (c) Looking from the back of the slide projector, the slide contains a letter ‘L’. What shape will appear on the screen? (d) The slide projector is moved closer to the screen. The image becomes unclear. Should the lens system be moved closer to or further away from the slide? 1 µm 82° light rays optical fibre 18. (a) You are carrying out a convex lens investigation at a bench near the classroom window and you obtain a sharp image of the window on your screen. A teacher walks past outside the window. What do you see on the screen? (b) The trees outside the classroom are unclear on the screen. What can you do to bring the trees into focus? 19. Use the ray tracing or the formula method to determine the magnification of an object placed under the following two-lens microscope. The object is placed 5.2 mm from an objective lens of focal length 5.0 mm. The eyepiece lens has a focal length of 40 mm. The poles of the lenses are 150 mm apart. 20. Light rays are shown passing through boxes in the figure at the base of the page. Identify the contents of each box from the options (a)–(g) given below. Option (b) is a mirror. All others are solid glass. Note : There are more options than boxes. 21. A convex lens with a focal length of 5.0 cm is used as a magnifying glass. Calculate the size and location of the image of text on this page if the centre of the lens was placed: (a) 4.0 cm above the page (b) 3.0 cm above the page. 23. A teacher is using a slide projector but the image on the screen is smaller than the screen. What needs to be done to produce a clear image on the full screen? 24. Calculate the distance the photographic film needs to be from the centre of a camera lens of focal length 5.0 cm in order to take a sharp photograph of a family group 15.0 m away. 25. (a) To appear invisible you need to become transparent. What must your refractive index be if your movement is not to be detected? (b) The retina of your eye is a light-absorbing screen. What does that imply about your own vision if you are to remain invisible? (Hint: If you are invisible all light passes through you.) More of a challenge 26. Calculate the angle of deviation at a glass–air interface for an angle of incidence of 65° and refractive index of glass of 1.55. 27. Calculate the sideways deflection as a ray of light goes through a parallel-sided plastic block (n = 1.4) with sides 5.0 cm apart, as in the figure below. 30° 22. A 35 mm slide is placed in a slide projector. A sharp image is produced on a screen 4.0 m away. The focal length of the lens system is 5.0 cm. (a) How far is the slide from the centre of the lens? (b) What is the size of the image? (i) (a) (ii) (b) (iii) (c) n = 1.4 (iv) (d) (v) (e) 5 cm (vi) (f) (g) CHAPTER 2 REFRACTING LIGHT 47 CHAPTER REVIEW time delay between the two rays after one internal reflection. Do you think this could be a problem in an optical fibre? If so, when? How could the problem be overcome? Jac Phys 1 2E - 02 Page 48 Tuesday, October 21, 2003 1:43 PM 28. Calculate the angle of deviation as the light ray goes through the triangular prism shown in the figure below. n = 1.5 60° ? 40° 29. A ray of light enters a glass sphere (n = 1.5), as in the figure below. What happens to the ray? glass sphere 30° CHAPTER REVIEW centre of circle 30. (a) Model a heat mirage on a spreadsheet (see the example at the base of the page) as a set of air layers whose refractive indices each decrease by 0.000 02, starting at 1.0003. As the light ray travels through each layer, the angle of refraction at the layer’s top surface becomes the angle of incidence at its bottom surface. Total internal reflection will occur when the spreadsheet tries to find an angle whose sine is greater than 1 and displays #NUM. (b) If the initial angle of incidence is 89°, in what layer does total internal reflection occur? Experiment with different initial angles of incidence, initial refractive indices and different rates of change in the refractive index from layer to layer. 31. In the diagram of Fata Morgana (figure 2.12, page 30), describe how the temperature and refractive index vary with height. 32. A hair 1.0 mm in diameter is looked at through a microscope. The objective lens has a focal length of 3.0 mm, the eyepiece has a 48 WAVE-LIKE PROPERTIES OF LIGHT focal length of 25 mm and the lenses are 18.0 cm apart. The hair is placed just outside the focus of the objective lens at 3.5 mm. (a) Calculate the location of the image produced by this lens. Using this image as the object for the eyepiece lens, calculate the location of the final image. (b) Calculate the size of the image produced by the objective lens, then calculate the size of the final image produced by the microscope. 33. A screen captures the image of a small globe from a convex lens. The screen is then taken away. Where should you place your eye and where should you look to see the image? 34. A hollow 15 cm tube contains a convex lens of unknown focal length but greater than 15 cm. However, you do not know where the lens is in the tube. You have a light source, a screen and a ruler. How would you determine the focal length of the lens and its position in the tube? 35. Ray tracing can be used to locate images produced by a concave lens. The only difference from ray tracing with convex lenses is that the principal focus is on the opposite side of the concave lens from the object. Draw ray tracing diagrams for objects at different distances and describe the properties of images produced by a concave lens. 36. The closer an object is to your eye, the more detail you can see. The object casts a larger visual angle at the eye. But this is only true up to your near point, which is the limit of your eye’s focusing ability. This is an important consideration in the design of magnifying glasses and microscopes. For the most detailed image, the device should produce an image located at the user’s near point. (a) A slide viewer is a small magnifying glass in a frame which holds a slide at a distance of 5.0 cm from the magnifying lens. What should be the focal length of the lens if the average near point is 25 cm and in normal use the slide viewer is held up to the eye? (b) A person with a near point of 50 cm asks you to design a magnifying glass to enable the person to read a book. What focal length convex lens should you use? (Remember to estimate object and image distances.)