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Chapter Nine
RAY OPTICS
AND OPTICAL
INSTRUMENTS
9.1 INTRODUCTION
Nature has endowed the human eye (retina) with the sensitivity to detect electromagnetic waves within a small range of the electromagnetic spectrum. Electromagnetic radiation belonging to this region of the spectrum (wavelength of about 400 nm to 750 nm) is called light. It is mainly through light and the sense of vision that we know and interpret the world around us. There are two things that we can intuitively mention about light from common experience. First, that it travels with enormous speed and second, that it travels in a straight line. It took some time for people to realise that the speed of light is finite and measurable. Its presently accepted value in vacuum is c = 2.99792458 × 10 8 m s–1. For many purposes, it suffices to take c = 3 × 10 8 m s –1^. The speed of light in vacuum is the highest speed attainable in nature. The intuitive notion that light travels in a straight line seems to contradict what we have learnt in Chapter 8, that light is an electromagnetic wave of wavelength belonging to the visible part of the spectrum. How to reconcile the two facts? The answer is that the wavelength of light is very small compared to the size of ordinary objects that we encounter commonly (generally of the order of a few cm or larger). In this situation, as you will learn in Chapter 10, a light wave can be considered to travel from one point to another, along a straight line joining
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them. The path is called a ray of light, and a bundle of such rays constitutes a beam of light. In this chapter, we consider the phenomena of reflection, refraction and dispersion of light, using the ray picture of light. Using the basic laws of reflection and refraction, we shall study the image formation by plane and spherical reflecting and refracting surfaces. We then go on to describe the construction and working of some important optical instruments, including the human eye.
PARTICLE MODEL OF LIGHT
Newton’s fundamental contributions to mathematics, mechanics, and gravitation often blind us to his deep experimental and theoretical study of light. He made pioneering contributions in the field of optics. He further developed the corpuscular model of light proposed by Descartes. It presumes that light energy is concentrated in tiny particles called corpuscles. He further assumed that corpuscles of light were massless elastic particles. With his understanding of mechanics, he could come up with a simple model of reflection and refraction. It is a common observation that a ball bouncing from a smooth plane surface obeys the laws of reflection. When this is an elastic collision, the magnitude of the velocity remains the same. As the surface is smooth, there is no force acting parallel to the surface, so the component of momentum in this direction also remains the same. Only the component perpendicular to the surface, i.e., the normal component of the momentum, gets reversed in reflection. Newton argued that smooth surfaces like mirrors reflect the corpuscles in a similar manner. In order to explain the phenomena of refraction, Newton postulated that the speed of the corpuscles was greater in water or glass than in air. However, later on it was discovered that the speed of light is less in water or glass than in air. In the field of optics, Newton – the experimenter, was greater than Newton – the theorist. He himself observed many phenomena, which were difficult to understand in terms of particle nature of light. For example, the colours observed due to a thin film of oil on water. Property of partial reflection of light is yet another such example. Everyone who has looked into the water in a pond sees image of the face in it, but also sees the bottom of the pond. Newton argued that some of the corpuscles, which fall on the water, get reflected and some get transmitted. But what property could distinguish these two kinds of corpuscles? Newton had to postulate some kind of unpredictable, chance phenomenon, which decided whether an individual corpuscle would be reflected or not. In explaining other phenomena, however, the corpuscles were presumed to behave as if they are identical. Such a dilemma does not occur in the wave picture of light. An incoming wave can be divided into two weaker waves at the boundary between air and water.
9.2 R EFLECTION OF LIGHT BY SPHERICAL MIRRORS
We are familiar with the laws of reflection. The angle of reflection (i.e., the angle between reflected ray and the normal to the reflecting surface or the mirror) equals the angle of incidence (angle between incident ray and the normal). Also that the incident ray, reflected ray and the normal to the reflecting surface at the point of incidence lie in the same plane (Fig. 9.1). These laws are valid at each point on any reflecting surface whether plane or curved. However, we shall restrict our discussion to the special case of curved surfaces, that is, spherical surfaces. The normal in
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The distance between the focus F and the pole P of the mirror is called the focal length of the mirror, denoted by f. We now show that f = R /2, where R is the radius of curvature of the mirror. The geometry of reflection of an incident ray is shown in Fig. 9.4. Let C be the centre of curvature of the mirror. Consider a ray parallel to the principal axis striking the mirror at M. Then CM will be perpendicular to the mirror at M. Let θ be the angle of incidence, and MD be the perpendicular from M on the principal axis. Then, ∠MCP = θ and ∠MFP = 2 θ Now,
tan θ =
MD CD and tan 2^ θ =^
MD FD (9.1) For small θ, which is true for paraxial rays, tan θ ≈θ, tan 2 θ ≈ 2 θ. Therefore, Eq. (9.1) gives MD FD =^2
MD CD
or, FD =
CD 2 (9.2) Now, for small θ, the point D is very close to the point P. Therefore, FD = f and CD = R. Equation (9.2) then gives f = R/ 2 (9.3)
9.2.3 The mirror equation
If rays emanating from a point actually meet at another point after reflection and/or refraction, that point is called the image of the first point. The image is real if the rays actually converge to the point; it is
FIGURE 9.3 Focus of a concave and convex mirror.
FIGURE 9.4 Geometry of reflection of an incident ray on (a) concave spherical mirror, and (b) convex spherical mirror.
Ray Optics and Optical Instruments
virtual if the rays do not actually meet but appear to diverge from the point when produced backwards. An image is thus a point-to-point correspondence with the object established through reflection and/or refraction. In principle, we can take any two rays emanating from a point on an object, trace their paths, find their point of intersection and thus, obtain the image of the point due to reflection at a spherical mirror. In practice, however, it is convenient to choose any two of the following rays: (i) The ray from the point which is parallel to the principal axis. The reflected ray goes through the focus of the mirror. (ii) The ray passing through the centre of curvature of a concave mirror or appearing to pass through it for a convex mirror. The reflected ray simply retraces the path. (iii) The ray passing through (or directed towards) the focus of the concave mirror or appearing to pass through (or directed towards) the focus of a convex mirror. The reflected ray is parallel to the principal axis. (iv) The ray incident at any angle at the pole. The reflected ray follows laws of reflection. Figure 9.5 shows the ray diagram considering three rays. It shows the image A′B′ (in this case, real) of an object A B formed by a concave mirror. It does not mean that only three rays emanate from the point A. An infinite number of rays emanate from any source, in all directions. Thus, point A′ is image point of A if every ray originating at point A and falling on the concave mirror after reflection passes through the point A′. We now derive the mirror equation or the relation between the object distance ( u ), image distance ( v ) and the focal length ( f ). From Fig. 9.5, the two right-angled triangles A′B′F and MPF are similar. (For paraxial rays, MP can be considered to be a straight line perpendicular to CP.) Therefore,
B A B F PM FP
or
B A B F BA FP
( PM = AB) (9.4)
Since ∠ APB = ∠ A′PB′, the right angled triangles A′B′P and ABP are also similar. Therefore,
B A B P B A B P (9.5) Comparing Eqs. (9.4) and (9.5), we get
B F B P – FP B P FP FP BP
Equation (9.6) is a relation involving magnitude of distances. We now apply the sign convention. We note that light travels from the object to the mirror MPN. Hence this is taken as the positive direction. To reach
FIGURE 9.5 Ray diagram for image formation by a concave mirror.
Ray Optics and Optical Instruments
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(^) 9.
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(^) 9.
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(^) 9.
Example 9.1 Suppose that the lower half of the concave mirror’s reflecting surface in Fig. 9.5 is covered with an opaque (non-reflective) material. What effect will this have on the image of an object placed in front of the mirror?
Solution You may think that the image will now show only half of the object, but taking the laws of reflection to be true for all points of the remaining part of the mirror, the image will be that of the whole object. However, as the area of the reflecting surface has been reduced, the intensity of the image will be low (in this case, half).
Example 9.2 A mobile phone lies along the principal axis of a concave mirror, as shown in Fig. 9.7. Show by suitable diagram, the formation of its image. Explain why the magnification is not uniform. Will the distortion of image depend on the location of the phone with respect to the mirror?
FIGURE 9. Solution The ray diagram for the formation of the image of the phone is shown in Fig. 9.7. The image of the part which is on the plane perpendicular to principal axis will be on the same plane. It will be of the same size, i.e., B′C = BC. You can yourself realise why the image is distorted.
Example 9.3 An object is placed at (i) 10 cm, (ii) 5 cm in front of a concave mirror of radius of curvature 15 cm. Find the position, nature, and magnification of the image in each case.
Solution
The focal length f = –15/2 cm = –7.5 cm (i) The object distance u = –10 cm. Then Eq. (9.7) gives
1 1 1 v 10 7 5
or
. .
10 7 5 2 5
v = (^) = – 30 cm
The image is 30 cm from the mirror on the same side as the object.
Also, magnification m =
v u The image is magnified, real and inverted.
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(ii) The object distance u = –5 cm. Then from Eq. (9.7), 1 1 1 v 5 7.
or
.
. –
5 7 5 15 cm 7 5 5
v =
This image is formed at 15 cm behind the mirror. It is a virtual image.
Magnification m = – – 15 3 ( 5)
v u The image is magnified, virtual and erect.
Example 9.4 Suppose while sitting in a parked car, you notice a jogger approaching towards you in the side view mirror of R = 2 m. If the jogger is running at a speed of 5 m s–1, how fast the image of the jogger appear to move when the jogger is (a) 39 m, (b) 29 m, (c) 19 m, and (d) 9 m away. Solution From the mirror equation, Eq. (9.7), we get fu v u f For convex mirror, since R = 2 m, f = 1 m. Then
for u = –39 m,
( 39) (^1 39) m 39 1 40
v
Since the jogger moves at a constant speed of 5 m s–1^ , after 1 s the position of the image v (for u = –39 + 5 = –34) is (34/35 )m. The shift in the position of image in 1 s is (^39 34 1365 13605 1) m 40 35 1400 1400 280 Therefore, the average speed of the image when the jogger is between 39 m and 34 m from the mirror, is (1/280) m s – Similarly, it can be seen that for u = –29 m, –19 m and –9 m, the speed with which the image appears to move is (^1) m s –1 (^) , 1 m s –1 (^) and 1 m s–1 , 150 60 10 respectively. Although the jogger has been moving with a constant speed, the speed of his/her image appears to increase substantially as he/she moves closer to the mirror. This phenomenon can be noticed by any person sitting in a stationary car or a bus. In case of moving vehicles, a similar phenomenon could be observed if the vehicle in the rear is moving closer with a constant speed.
9.3 R EFRACTION
When a beam of light encounters another transparent medium, a part of light gets reflected back into the first medium while the rest enters the other. A ray of light represents a beam. The direction of propagation of an obliquely incident ray of light that enters the other medium, changes
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deviation, but it does suffer lateral displacement/ shift with respect to the incident ray. Another familiar observation is that the bottom of a tank filled with water appears to be raised (Fig. 9.10). For viewing near the normal direction, it can be shown that the apparent depth, ( h 1 ) is real depth ( h (^) 2 ) divided by the refractive index of the medium (water). The refraction of light through the atmosphere is responsible for many interesting phenomena. For example, the sun is visible a little before the actual sunrise and until a little after the actual sunset due to refraction of light through the atmosphere (Fig. 9.11). By actual sunrise we mean the actual crossing of the horizon by the sun. Figure 9. shows the actual and apparent positions of the sun with respect to the horizon. The figure is highly exaggerated to show the effect. The refractive index of air with respect to vacuum is 1.00029. Due to this, the apparent shift in the direction of the sun is by about half a degree and the corresponding time difference between actual sunset and apparent sunset is about 2 minutes (see Example 9.5). The apparent flattening (oval shape) of the sun at sunset and sunrise is also due to the same phenomenon.
FIGURE 9.10 Apparent depth for (a) normal, and (b) oblique viewing.
FIGURE 9.11 Advance sunrise and delayed sunset due to atmospheric refraction.
Example 9.5 The earth takes 24 h to rotate once about its axis. How much time does the sun take to shift by 1º when viewed from the earth? Solution Time taken for 360° shift = 24 h Time taken for 1° shift = 24/360 h = 4 min.
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9.4 T OTAL INTERNAL R EFLECTION
When light travels from an optically denser medium to a rarer medium at the interface, it is partly reflected back into the same medium and partly refracted to the second medium. This reflection is called the internal reflection. When a ray of light enters from a denser medium to a rarer medium, it bends away from the normal, for example, the ray AO 1 B in Fig. 9.12. The incident ray AO 1 is partially reflected (O 1 C) and partially transmitted (O 1 B) or refracted, the angle of refraction ( r ) being larger than the angle of incidence ( i ). As the angle of incidence increases, so does the angle of refraction, till for the ray AO 3 , the angle of refraction is π/2. The refracted ray is bent so much away from the normal that it grazes the surface at the interface between the two media. This is shown by the ray AO 3 D in Fig. 9.12. If the angle of incidence is increased still further (e.g., the ray AO 4 ), refraction is not possible, and the incident ray is totally reflected.
THE DROWNING CHILD, LIFEGUARD AND SNELL’S LAW
Consider a rectangular swimming pool PQSR; see figure here. A lifeguard sitting at G outside the pool notices a child drowning at a point C. The guard wants to reach the child in the shortest possible time. Let SR be the side of the pool between G and C. Should he/she take a straight line path GAC between G and C or GBC in which the path BC in water would be the shortest, or some other path GXC? The guard knows that his/her running speed v 1 on ground is higher than his/her swimming speed v 2. Suppose the guard enters water at X. Let GX = l 1 and XC = l (^) 2. Then the time taken to reach from G to C would be
1 2 1 2
l l t v v To make this time minimum, one has to differentiate it (with respect to the coordinate of X ) and find the point X when t is a minimum. On doing all this algebra (which we skip here), we find that the guard should enter water at a point where Snell’s law is satisfied. To understand this, draw a perpendicular LM to side SR at X. Let ∠GXM = i and ∠CXL = r. Then it can be seen that t is minimum when
1 2
sin sin
i v r v In the case of light v 1 /v 2 , the ratio of the velocity of light in vacuum to that in the medium, is the refractive index n of the medium. In short, whether it is a wave or a particle or a human being, whenever two mediums and two velocities are involved, one must follow Snell’s law if one wants to take the shortest time.
Ray Optics and Optical Instruments
above the water surface is totally absent and the beam is totally reflected back to water. This is total internal reflection at its simplest. Pour this water in a long test tube and shine the laser light from top, as shown in Fig. 9.13(c). Adjust the direction of the laser beam such that it is totally internally reflected every time it strikes the walls of the tube. This is similar to what happens in optical fibres. Take care not to look into the laser beam directly and not to point it at anybody’s face.
9.4.1 Total internal reflection in nature and
its technological applications
(i) Mirage: On hot summer days, the air near the ground becomes hotter than the air at higher levels. The refractive index of air increases with its density. Hotter air is less dense, and has smaller refractive index than the cooler air. If the air currents are small, that is, the air is still, the optical density at different layers of air increases with height. As a result, light from a tall object such as a tree, passes through a medium whose refractive index decreases towards the ground. Thus, a ray of light from such an object successively bends away from the normal and undergoes total internal reflection, if the angle of incidence for the air near the ground exceeds the critical angle. This is shown in Fig. 9.14(b). To a distant observer, the light appears to be coming from somewhere below the ground. The observer naturally assumes that light is being reflected from the ground, say, by a pool of water near the tall object. Such inverted images of distant tall objects cause an optical illusion to the observer. This phenomenon is called mirage. This type of mirage is especially common in hot deserts. Some of you might have noticed that while moving in a bus or a car during a hot summer day, a distant patch of road, especially on a highway, appears to be wet. But, you do not find any evidence of wetness when you reach that spot. This is also due to mirage.
FIGURE 9. Observing total internal reflection in water with a laser beam (refraction due to glass of beaker neglected being very thin).
FIGURE 9.14 (a) A tree is seen by an observer at its place when the air above the ground is at uniform temperature, (b) When the layers of air close to the ground have varying temperature with hottest layers near the ground, light from a distant tree may undergo total internal reflection, and the apparent image of the tree may create an illusion to the observer that the tree is near a pool of water.
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(ii) Diamond : Diamonds are known for their spectacular brilliance. Their brilliance is mainly due to the total internal reflection of light inside them. The critical angle for diamond-air interface (≅ 24.4°) is very small, therefore once light enters a diamond, it is very likely to undergo total internal reflection inside it. Diamonds found in nature rarely exhibit the brilliance for which they are known. It is the technical skill of a diamond cutter which makes diamonds to sparkle so brilliantly. By cutting the diamond suitably, multiple total internal reflections can be made to occur. (iii) Prism : Prisms designed to bend light by 90º or by 180º make use of total internal reflection [Fig. 9.15(a) and (b)]. Such a prism is also used to invert images without changing their size [Fig. 9.15(c)]. In the first two cases, the critical angle i (^) c for the material of the prism must be less than 45º. We see from Table 9.1 that this is true for both crown glass and dense flint glass. (iv) Optical fibres : Now-a-days optical fibres are extensively used for transmitting audio and video signals through long distances. Optical fibres too make use of the phenomenon of total internal reflection. Optical fibres are fabricated with high quality composite glass/quartz fibres. Each fibre consists of a core and cladding. The refractive index of the material of the core is higher than that of the cladding. When a signal in the form of light is directed at one end of the fibre at a suitable angle, it undergoes repeated total internal reflections along the length of the fibre and finally comes out at the other end (Fig. 9.16). Since light undergoes total internal reflection at each stage, there is no appreciable loss in the intensity of the light signal. Optical fibres are fabricated such that light reflected at one side of inner surface strikes the other at an angle larger than the critical angle. Even if the fibre is bent, light can easily travel along its length. Thus, an optical fibre can be used to act as an optical pipe. A bundle of optical fibres can be put to several uses. Optical fibres are extensively used for transmitting and receiving electrical signals which are converted to light by suitable transducers. Obviously, optical fibres can also be used for transmission of optical signals. For example, these are used as a ‘light pipe’ to facilitate visual examination of internal organs like esophagus, stomach and intestines. You might have seen a commonly
FIGURE 9.15 Prisms designed to bend rays by 90º and 180º or to invert image without changing its size make use of total internal reflection.
FIGURE 9.16 Light undergoes successive total internal reflections as it moves through an optical fibre.
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Now, for ΔNOC, i is the exterior angle. Therefore, i = ∠NOM + ∠NCM
i = MN^ MN OM MC
Similarly, r = ∠NCM – ∠NIM
i.e., r =
MN MN MC MI (9.14)
Now, by Snell’s law n 1 sin i = n (^) 2 sin r or for small angles n 1 i = n (^) 2 r
LIGHT SOURCES AND PHOTOMETRY
It is known that a body above absolute zero temperature emits electromagnetic radiation. The wavelength region in which the body emits the radiation depends on its absolute temperature. Radiation emitted by a hot body, for example, a tungsten filament lamp having temperature 2850 K are partly invisible and mostly in infrared (or heat) region. As the temperature of the body increases radiation emitted by it is in visible region. The sun with temperature of about 5500 K emits radiation whose energy versus wavelength graph peaks approximately at 550 nm corresponding to green light and is almost in the middle of the visible region. The energy versus wavelength distribution graph for a given body peaks at some wavelength, which is inversely proportional to the absolute temperature of that body. The measurement of light as perceived by human eye is called photometry. Photometry is measurement of a physiological phenomenon, being the stimulus of light as received by the human eye, transmitted by the optic nerves and analysed by the brain. The main physical quantities in photometry are (i) the luminous intensity of the source, (ii) the luminous flux or flow of light from the source, and (iii) illuminance of the surface. The SI unit of luminous intensity ( I ) is candela (cd). The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540 × 10^12 Hz and that has a radiant intensity in that direction of 1/683 watt per steradian. If a light source emits one candela of luminous intensity into a solid angle of one steradian, the total luminous flux emitted into that solid angle is one lumen (lm). A standard 100 watt incadescent light bulb emits approximately 1700 lumens. In photometry, the only parameter, which can be measured directly is illuminance. It is defined as luminous flux incident per unit area on a surface (lm/m^2 or lux ). Most light meters measure this quantity. The illuminance E , produced by a source of luminous intensity I , is given by E = I / r^2 , where r is the normal distance of the surface from the source. A quantity named luminance ( L ), is used to characterise the brightness of emitting or reflecting flat surfaces. Its unit is cd/m^2 (sometimes called ‘nit’ in industry). A good LCD computer monitor has a brightness of about 250 nits.
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(^) 9.
Substituting i and r from Eqs. (9.13) and (9.14), we get
1 2 2 1 OM MI MC
n n n n (9.15)
Here, OM, MI and MC represent magnitudes of distances. Applying the Cartesian sign convention,
OM = – u , MI = + v , MC = + R
Substituting these in Eq. (9.15), we get
n (^) 2 n (^) 1 n (^) 2 n 1 v u R (9.16) Equation (9.16) gives us a relation between object and image distance in terms of refractive index of the medium and the radius of curvature of the curved spherical surface. It holds for any curved spherical surface.
Example 9.6 Light from a point source in air falls on a spherical glass surface ( n = 1.5 and radius of curvature = 20 cm). The distance of the light source from the glass surface is 100 cm. At what position the image is formed? Solution We use the relation given by Eq. (9.16). Here u = – 100 cm, v = ?, R = + 20 cm , n 1 = 1, and n 2 = 1.5. We then have 1.5 1 0. v 100 20 or v = +100 cm The image is formed at a distance of 100 cm from the glass surface, in the direction of incident light.
9.5.2 Refraction by a lens
Figure 9.18(a) shows the geometry of image formation by a double convex lens. The image formation can be seen in terms of two steps: (i) The first refracting surface forms the image I 1 of the object O [Fig. 9.18(b)]. The image I 1 acts as a virtual object for the second surface that forms the image at I [Fig. 9.18(c)]. Applying Eq. (9.15) to the first interface ABC, we get
1 2 2 1 OB BI 1 BC 1
n n n n (9.17)
A similar procedure applied to the second interface* ADC gives,
2 1 2 1 DI 1 DI DC 2
n n n n (9.18)
- Note that now the refractive index of the medium on the right side of ADC is n 1 while on its left it is n 2. Further DI 1 is negative as the distance is measured against the direction of incident light.
Ray Optics and Optical Instruments
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(^) 9.
the laws of refraction and find the point where the refracted rays meet (or appear to meet). In practice, however, it is convenient to choose any two of the following rays: (i) A ray emanating from the object parallel to the principal axis of the lens after refraction passes through the second principal focus F′ (in a convex lens) or appears to diverge (in a concave lens) from the first principal focus F. (ii) A ray of light, passing through the optical centre of the lens, emerges without any deviation after refraction. (iii) A ray of light passing through the first principal focus (for a convex lens) or appearing to meet at it (for a concave lens) emerges parallel to the principal axis after refraction. Figures 9.19(a) and (b) illustrate these rules for a convex and a concave lens, respectively. You should practice drawing similar ray diagrams for different positions of the object with respect to the lens and also verify that the lens formula, Eq. (9.23), holds good for all cases. Here again it must be remembered that each point on an object gives out infinite number of rays. All these rays will pass through the same image point after refraction at the lens. Magnification ( m ) produced by a lens is defined, like that for a mirror, as the ratio of the size of the image to that of the object. Proceeding in the same way as for spherical mirrors, it is easily seen that for a lens
m =
h h =^
v u (9.24) When we apply the sign convention, we see that, for erect (and virtual) image formed by a convex or concave lens, m is positive, while for an inverted (and real) image, m is negative.
Example 9.7 A magician during a show makes a glass lens with n = 1.47 disappear in a trough of liquid. What is the refractive index of the liquid? Could the liquid be water? Solution The refractive index of the liquid must be equal to 1.47 in order to make the lens disappear. This means n 1 = n 2.. This gives 1/ f = 0 or f → ∞. The lens in the liquid will act like a plane sheet of glass. No, the liquid is not water. It could be glycerine.
9.5.3 Power of a lens
Power of a lens is a measure of the convergence or divergence, which a lens introduces in the light falling on it. Clearly, a lens of shorter focal
FIGURE 9.19 Tracing rays through (a) convex lens (b) concave lens.
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length bends the incident light more, while converging it in case of a convex lens and diverging it in case of a concave lens. The power P of a lens is defined as the tangent of the angle by which it converges or diverges a beam of light falling at unit distant from the optical centre (Fig. 9.20).
tan^ h^ ; if h 1 tan^1 f f or^
f for small value of δ. Thus,
P =
1 f (9.25) The SI unit for power of a lens is dioptre (D): 1D = 1m–1^. The power of a lens of focal length of 1 metre is one dioptre. Power of a lens is positive for a converging lens and negative for a diverging lens. Thus, when an optician prescribes a corrective lens of power + 2.5 D, the required lens is a convex lens of focal length + 40 cm. A lens of power of – 4.0 D means a concave lens of focal length – 25 cm.
Example 9.8 (i) If f = 0.5 m for a glass lens, what is the power of the lens? (ii) The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. Its focal length is 12 cm. What is the refractive index of glass? (iii) A convex lens has 20 cm focal length in air. What is focal length in water? (Refractive index of air-water = 1.33, refractive index for air-glass = 1.5.) Solution (i) Power = +2 dioptre. (ii) Here, we have f = +12 cm, R 1 = +10 cm, R 2 = –15 cm. Refractive index of air is taken as unity. We use the lens formula of Eq. (9.22). The sign convention has to be applied for f , R 1 and R 2. Substituting the values, we have 1 1 1 ( 1) 12 10 15
n This gives n = 1.5. (iii) For a glass lens in air, n 2 = 1.5, n 1 = 1, f = +20 cm. Hence, the lens formula gives
1 2
(^1) 0.5 1 1 20 R R For the same glass lens in water, n 2 = 1.5, n 1 = 1.33. Therefore,
1 2
1.33 (^) (1.5 1.33) 1 1 f R R (9.26) Combining these two equations, we find f = + 78.2 cm.
9.5.4 Combination of thin lenses in contact
Consider two lenses A and B of focal length f 1 and f 2 placed in contact with each other. Let the object be placed at a point O beyond the focus of
FIGURE 9.20 Power of a lens.
