OPTICS: Solutions to higher level questions

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1 OPTICS: Solutions to higher level questions 2015 Question 12 (b) (i) Complete the path of the light ray through the section of the lens. See diagram: (ii) Draw a ray diagram to show the formation of a virtual image in a converging lens. See diagram: (iii)a converging lens of focal length 20 cm and a diverging lens of focal length 8 cm are placed in contact. Calculate the power of the combination. P1 = 1/0.2 and P2 = 1/ 0.08 P = P1 + P2 P = 7.5 m 1 (iv) What eye defect can be corrected using converging lenses? Long-sightedness 2014 Question 12 (b) (i) What is reflection? light rebounding (off surfaces) (ii) Draw a ray diagram to show the formation of an image in a convex mirror. object, correct mirror two correct reflected rays correct image (iii)find the position of the image. f = r/2 = 5 cm 1/u + 1/v = 1/f v = 30/7 = 4.3 cm (behind the mirror) (iv) Concave mirrors, rather than convex mirrors, are used by dentists to examine teeth. Explain why. to give a magnified image 1

2 2012 Question 12 (b) (i) If the refractive index of the glass is 1.5, calculate the value of θ. This is one of the nastiest questions I have seen on any paper. First, recognise that that θ does not represent the angle of incidence; you need to subtract the angle of incidence from 90 to get θ. Secondly, the angle of refraction r is 60 0 not Finally, anytime you are told that the refractive index of a material is 1.5, what is implied is that light is going from air into this medium (go back and learn the full definition of refractive index to see why); in this case the light is going from the medium (glass) to air, therefore the refractive index is sin i = 1 sin Therefore i = Therefore θ = Told you it was nasty (ii) What would be the value of the angle θ so that the ray of light emerges parallel to the side of the glass block? 1 n = sin c Here we don t have to worry about the direction of light; the formula for the critical angle assumes that light is going from the medium to air. c = θ = (iii)calculate the speed of light as it passes through the glass. n g = c a c g c g = 1.5 = m s Question 12 (b) (i) State the laws of refraction of light. The incident ray, refracted ray and normal all lie in same plane sin i/sinr is a constant (ii) Draw a ray diagram to show where the lamp appears to be, as seen by an observer standing at the edge of the pool. (iii)explain why the area of water surrounding the disc of light appears dark. No light emerges from this area of the pool due to total internal reflection (iv) Calculate the area of the illuminated disc of water. η = 1 sin i c η = 1.33 ic = radius of disc = r = 1.8 tan(48.76) r = m area = πr 2 = m 2 2

3 2009 Question 12 (c) (i) Explain, with the aid of a labelled diagram, how a ray of light is guided along a fibre. 1. An optical fibre consists of a glass pipe coated with a second material of lower refractive index. 2. Light enters one end of the fibre and strikes the boundary between the two materials at an angle greater than the critical angle, resulting in total internal reflection at the interface. 3. This reflected light now strikes the interface on the opposite wall and gets totally reflected again. 4. This process continues all along the glass pipe until the light emerges at the far end. (ii) Why is each fibre coated with glass of lower refractive index? Because total internal reflection can only occur for rays travelling from a denser to a rarer medium. (iii)what is the speed of the light as it passes through the fibre? n = cair/cglass glass = /1.55 cglass = m s -1 (iv) What is the power being transmitted by the light after it has travelled 8 km through the fibre? After 2 km power has dropped to 5 W; after 4 km power has dropped to 2.5 W; after 6 km power has dropped to 1.25 W; after 8 km power has dropped to W Question 9 (i) What is meant by refraction of light? Refraction is the bending of light as it passes from one medium to another (of different refractive index). (ii) State Snell s law of refraction. The ratio of the sin of the sin of the angle of incidence to the sin of the angle of refraction is a constant. (iii)calculate how near an object can be placed in front of the eye and still be in focus. Pmax = 64 m -1 = 1/f f = m = 1.56 cm 1/u = 1/v +1/f (iv) Calculate the maximum power of the internal lens. Pmax = P1 + P2 2 2 = 26 m -1 (v) Light is refracted as it enters the cornea from air as shown in the diagram. Calculate the refractive index of the cornea. sin i = n sin r n = sin 37/sin 27 n = 1.33 (vi) Draw a diagram to show the path of a ray of light as it passes from water of refractive index 1.33 into the cornea. Both media have the same refractive index so there is no bending of light. Draw a straight line passing from one medium to the other without bending. (vii) When underwater why does the cornea not act as a lens? Because light does not refract at the cornea since there is no change in refractive index. (viii) What is the maximum power of the eye when underwater? The maximum power of the eye is 64 m 1, but this includes the focusing power of the cornea (38 m -1 ) which doesn t work underwater, so maximum power = = 26 m 1. (ix) Why do objects appear blurred when underwater? Because the internal lens by itself is not powerful enough to focus light on retina. (x) Explain how wearing goggles allows objects to be seen clearly. Because light which hits the cornea is coming from air and so there will be refraction here (the cornea will now act as a lens). 3

4 2006 Question 7 (i) What is meant by the refraction of light? Refraction is the bending of light as it passes from one medium to another. (ii) A converging lens is used as a magnifying glass. Draw a ray diagram to show how an erect image is formed by a magnifying glass. Object inside focal point Two (appropriate) rays from object to lens Two rays emerge correctly from lens Rays produced back to form upright virtual image (on same side as object) (iii)a diverging lens cannot be used as a magnifying glass. Explain why. The image is always dimished. (iv) The converging lens has a focal length of 8 cm. Determine the two positions that an object can be placed to produce an image that is four times the size of the object? 1/u + 1/v = 1/f Magnification = v /u = 4 For real image: 1/u + 1/4u = 1/8 u = 10 cm For virtual image: 1 /u - 1/4u = 1/8 u = 6 cm (v) The power of an eye when looking at a distant object should be 60 m 1. A person with defective vision has a minimum power of 64 m 1. Calculate the focal length of the lens required to correct this defect. P = P1 + P2 P1 P1 = -4 ( m -1 ) P = 1/f f = 1/P f = (-)4 f = (-)¼ m = (-)25 cm (vi) What type of lens is used? Diverging / concave lens (vii) Name the defect. Short sight / myopia 2004 Question 12 (b) (i) Give two reasons why the telecommunications industry uses optical fibres instead of copper conductors to transmit signals. Less interference, boosted less often, cheaper raw material, occupy less space, more information carried in the same space, flexible for inaccessible places, do not corrode, etc (ii) Explain how a signal is transmitted along an optical fibre. Light ray introduced at one end of fibre and strikes the interface at an angle greater than the internal angle so total internal reflection occurs. This continues all along the fibre. (iii)an optical fibre has an outer less dense layer of glass. What is the role of this layer of glass? Total internal reflection will only occur if the outer medium is of greater density. It also prevents damage to the surface of the core. (iv) An optical fibre is manufactured using glass of refractive index of 1.5. Calculate the speed of light travelling through the optical fibre. ng = ca /cg 1.5 = / vg vg = (m s -1 ) 4

5 2002 Question 12 (b) (i) State the laws of refraction of light. The incident ray, the normal and the refracted ray all lie on the same plane. The ratio of the sin of the sin of the angle of incidence to the sin of the angle of refraction is a constant. (ii) Draw a labelled diagram showing the optical structure of the eye. See diagram. (iii)how does the eye bring objects at different distances into focus? It can change the shape of the lens which in turn changes the focal length of the lens. (iv) The power of a normal eye is +60 m -1. A short-sighted person s eye has a power of +65 m -1. Calculate the power of the contact lens required to correct the person s short-sightedness. PTotal = P1 + P2 60 = P Power = - 5 m (v) Calculate the focal length of the contact lens required to correct the person s short-sightedness m 5