FLAP P6.2 Rays and geometrical optics COPYRIGHT 1998 THE OPEN UNIVERSITY S570 V1.1

Transcription

1 F1 The ray approximation in optics assumes that light travels from one point to another along a narrow path called a ray that may be represented by a directed line (i.e. a line with an arrow on it). In a uniform medium (where the refractive index is the same everywhere) the rays are straight lines, though they may be kinked at mirror surfaces or upon passing through the interface between two different media and they may bend continuously in non-uniform media where the refractive index changes smoothly with position. The ray approximation gives an adequate description of light transmission provided diffraction effects can be ignored. This happens when the transverse dimensions of the optical systems or obstacles through or around which the light is passing are large compared with the wavelength of light.

2 F2 P normal P' A ray from the object to the eye is constructed by drawing the incident and reflected parts of the ray in the same plane as the normal at the point of incidence, and by making the angles of incidence and reflection equal (see Figure 3). From the geometry of the (similar) triangles shown in Figure 24, we can see that the virtual image will be 201cm behind the mirror and located along the extended normal from the object to the mirror. θ i A θ R mirror Figure 34A ray of light reflected from a plane mirror. O N θ i θ i θ R A mirror I' Figure 244See Answer T2.

3 F3 (a) We use Snell s law and assume the refractive index of air is 1.00: sin θ = sin θ therefore sin θ 1 = 0.667sin θ 2 The maximum value of sin1θ 2 is 1.00 and at this limiting value θ 1c = arcsin (0.667) = 41.8 This is the critical angle. When θ 1c > 41.8 there can be no refracted ray and total internal reflection occurs. (b) Light rays are confined in optical fibres by total internal reflection. These devices are used extensively in modern optical communication systems. See also Subsections 4.3 and 4.4.

4 Study comment Having seen the Fast track questions you may feel that it would be wiser to follow the normal route through the module and to proceed directly to Ready to study? in Subsection 1.3. Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly to the Closing items. If you have completed both the Fast track questions and the Exit test, then you have finished the module and may leave it here.

5 R1 If sin1θ = 0.45, then θ = arcsin1(0.45). To three significant figures this gives: θ = Consult trigonometry in the Glossary for further information.

6 R2 The relation between wavelength λ, speed c and frequency f of a wave is c = f0λ. The values for blue and red light are as follows: blue light λ = 4501nm, therefore f = Hz red light λ = 6501nm, therefore f = Hz Consult the relevant terms in the Glossary for further information.

7 R3 The rise and fall of a wave creates a vibration or oscillation at any point through which the wave passes. The term phase is generally used to describe the stage that an oscillator has reached in its repetitive oscillatory cycle, and is often expressed as a fraction of 2π since 2π radians represents one complete cycle. Two waves are said to be in phase at a certain point if the oscillations they would individually cause at that point are exactly in step, rising and falling together with equal phases at each moment. Under appropriate conditions, two waves arriving simultaneously at a given point may cause a joint disturbance at that point which is equal to the sum of the disturbances they would cause individually; under these circumstances the waves are said to interfere when they meet. Consult the relevant terms in the Glossary for further information if required, but few other wave concepts are needed in this module.

8 T1 See Figure 23. At the first reflection, point A, the reflected angle θ R1 is equal to the incident angle θ i1. The mirrors are at right angles so that at the second reflection, point B, θ i2 = 90 θ R1 and θ R2 = θ i2 = 90 θ R1. The angle of the ray relative to the horizontal after the second reflection is given by: α = 90 θ R2 = 90 (90 θ R1 ) = θ R1 = θ i1. The ray is finally reflected back parallel but the direction of travel is reversed. This is the basic principle of operation for a bicycle reflector and also for the reflectors placed on the Moon by the Apollo 11 astronauts (used to reflect laser beams sent from Earth and thereby to measure the distance from the Earth to the Moon1 1the lunar ranging experiment). A θ i1 θ R1 θ i2 θr2 α B Figure 234See Answer T1. mirror

9 T2 See Figure 24. Consider two rays from O1 1one along the normal to the mirror and the other at any angle of incidence θ1 i. The ray ON is reflected back towards O so the virtual image must lie somewhere along the normal. For the ray reflected at the arbitrary point A, the angle of incidence equals the angle of reflection: θ0 i = θ R. Hence the angle NI A = θ R. But the angle NOA = θ0 i = θ R so that the triangles ONA and I NA are identical. This means that side I N is equal to side ON for any value of θ i. All rays reflected in the mirror from O appear to diverge from I with I N = ON, so the image appears to be at the same distance from the mirror as the object. O N θ i θ i θ R A mirror I' Figure 244See Answer T2.

10 T3 The ray diagram for refraction through a parallel-sided glass block is shown in Figure 25. At A (the point where the light enters the block): µ 1 µ 2 µ 1 θ 2B B θ 1B µ 1 µ 2 = sin θ 2A sin θ 1A At B (the point where the light leaves the block): θ 1A A θ 2A µ 2 µ 1 = sin θ 1B sin θ 2B Multiplying these two equations together, and noting that θ 2A = θ 2B gives us Figure 254See Answer T3. 1 = sin θ 2A sin θ 1B = sin θ 1B sin θ 1A sin θ 2B sin θ 1A i.e. sin1θ 1B = sin1θ 1A and θ 1B = θ 1A ; the ray emerging from the block is parallel to the ray entering the block.

11 T4 We use the data given in Figure 8 for the variation of refractive index of a form of glass called fused quartz. Snell s law gives us sin θ air = µ glass sin θ glass µ air and we put µ air = At λ = 6501nm, µ glass = 1.46, sin 40 hence sin θ glass = = and θ glass = µ λ/nm Figure 84The variation of the refractive index with wavelength for fused quartz glass. (The wavelength is measured in a vacuum). At λ = 4501nm, µ glass = 1.465, sin 40 hence sin θ glass = = and θ glass = The difference in angles is = 0.10

12 T5 Using Snell s law for the glass air interface: µ air1 sin1θ air = µ glass 1sin1θ glass We can set µ air = 1.00 and then find: µ glass = sin1(35 )/sin1(22 ) = 0.574/0.375 = The refractive index of a material is defined as the ratio of the speed of light in a vacuum to the speed of light in the material, i.e. µ glass = c/v glass, so we have: v glass = c/µ glass = ( /1.53)1m1s 1 = m1s 1

13 T6 The real and apparent depth due to refraction are shown in Figure 26. If we take one ray normal to the interface along the line ON and another ray at angle θ 1 along OA. Snell s law gives us: sin θ 1 = µ 2 sin θ 2 µ 1 But when θ 1 and θ 2 are small: sin1θ 1 tan1θ 1 and sin1θ 2 tan1θ 2 N θ 1 A µ 1 µ 2 so that tan θ 1 = µ 2 tan θ 2 µ 1 therefore NA IN NA ON = µ 2 µ 1 thus ON IN 2 µ 1 as required in Equation 3 real depth apparent depth = ON IN 2 µ 1 (Eqn 3) I θ 1 θ 2 θ 2 O Figure 264See Answer T6.

14 From Equation 3 real depth apparent depth = ON IN =µ 2 (Eqn 3) µ 1 we have: real depth apparent depth = µ water = 1.33 µ air i.e. real depth = ( )1m = 1.331m

15 T7 The critical angle is given by Equation 5: θ c = arcsin µ 2 (Eqn 5) µ 1 θ c = arcsin µ water µ = arcsin 1.33 glass = 62.5 At incident angles greater than 62.5 the ray is totally internally reflected. Replacing the water by air gives θ c = arcsin1(1.00/1.50) = 41.8.

16 T8 Since it is continually entering layers of lower and lower refractive index it must be refracted smoothly and progressively further and further from the normal to the Earth s surface. The ray will follow a smooth curve and eventually will be bent back towards the surface, following a symmetric path and arriving at the surface some distance from its departure point. The atmosphere here acts like a mirror.

17 T9 Since the light ray is travelling along the direction of the changing refractive index it is always travelling along the normal, with an angle of incidence equal to 90, and so is not refracted at all1 1although its speed through the medium will change with the refractive index.

18 T10 See Figure 20. If the angles of incidence and reflection are to be equal then, with NA = x0: x a = d x b i.e. xb = ad ax Gathering terms involving x on the left-hand side, x(a + b) = ad so x = ad a + b Putting in the values for a, b and d we find 10 cm 20 cm x = (10 + 7)cm = 11.8cm P a θ i θ R b x N d A N P mirror Figure 204A conceivable route for light rays passing from P to P via a reflection at A; but is it the route the rays will actually follow? A careful drawing, using this value of x, gives a total true path length of 26.31mm. Nearby paths, with NA = 9.81cm or 13.81cm, are about 1.331mm longer, confirming that the true path is a minimum path.