CHAPTER TWO...REFRACTION AND REFLECTION

Transcription

1 The laws of refraction and reflection were discovered experimentally long before their significance was understood, and together they form the basis of the whole of geometrical optics. They may be derived from certain general principles to be discussed later, but for the present we shall merely state them as experimental facts. 2.1 Refraction Definition is defined as the changing of direction of a light ray and its speed of propagation as it passes from one medium into another. Light ray refracted in material, amount of refraction depends on refractive index (n) of material. Laws of refraction : - The refracted ray lies in the plane of incidence. - The sine of the angle of refraction bears a constant ratio to the sine of the angle of incidence. so that: ssssss ii ssssss rr constant In the refraction at a boundary between two such substances having incidences of refraction n 1 and n 2, Snell s law may be written in the form sin ii sin rr nn 2 nn 1 cccccccccccccccc or Snell s law n1 sin i n2 sin r Where: i: angle of incidence r: angle of refraction n 1 : refractive index of the medium 1 (Medium containing the incident ray) n 2 : refractive index of the medium 2 (Medium containing the refracted ray) Dr. Anees Ali Page 1

2 Figures 1 and 2 show the refraction of light travels from one medium to another medium (1) n 1 < n 2 (2) n 1 > n 2 Medium 1 is less density than medium 2 Medium 1 is denser than medium 2 The light ray is bent toward the normal, thus r < i The light ray is bent away the normal, thus r > i Notes: Depending on the Snell s law the refractive index can be defined as the constant ratio sin i / sin r for the two given media. - The value of refractive index depends on the type of medium and the colour of the light. It is dimensionless and its value > 1. - Consider the light ray travels from medium 1 into medium 2, the refractive index can be denoted by: 1n2 vvvvvvvvvvvvvvvv oooo llllllhtt iiii mmmmmmmmmmmm 1 vv1 vvvvvvvvvvvvvvvv oooo llllllhtt iiii mmmmmmmmmmmm 2 vv2 (Medium containing the incident ray) (Medium containing the refracted ray) - Absolute refractive index, n (for the incident ray is travelling in vacuum or air and is then refracted into the medium concerned is written by: nn cc vv Dr. Anees Ali Page 2

3 The relationship between refractive index and the wave length of light. As light travels from one medium to another, its wavelength, λ changes but its frequency, ƒ remains constant. The wavelength changes because of different materials while the frequency remains constant because the number of wave cycles arriving per unit time must equal the number leaving per unit time so that the boundary surface cannot create or destroy waves. By considering of light travels from medium 1 (n 1 ) into medium 2 (n 2 ), the velocity of light in each medium is given by: then v 1 ƒ λ 1 and v 2 ƒ λ 2 vv 1 vv 2 ƒ λλ 1 ƒ λλ 2 where v1 cc nn 1 and v2 cc nn 2 cc nn1 cc λλ 1 nn2 λλ 2 n1 λ1 n2 λ2 (Refractive index is inversely proportional to the wavelength) If medium 1 is vacuum or air, then n1 1. Hence the refractive index for any medium, n can be expressed as: Where: λ o : wavelength of light in vacuum λ : wavelength of light in medium n λλ oo λλ Dr. Anees Ali Page 3

4 Example 1: The speed of light in an unknown medium is measured to be 2.76 x 10 8 m/s. What is the refractive index of the medium? Solution: n c/v n (3.00 x 10 8 m/s) / (2.76 x 10 8 m/s) 1.09 Example 2: The wavelength of green light in air is 524 nm. If the light is travailing through the water (n 1.33). What is the (a) velocity, (b) wavelength and (c) the frequency of this light? Solution: (a) n c / v v c / n v ( m/s) / m/s. (b) n λ o / λ λ λ o / n λ 524 / nm. (c) v λ ƒ ƒ v / λ ƒ ( m/s) / m Hz. Remember when the light travels from one medium to another its wavelength, λ changes but its frequency, ƒ remains constant. So the frequency of the light in air equal to the frequency of light in water. The frequency of light in air ƒ c / λ o ƒ ( m/s) / m Hz. Dr. Anees Ali Page 4

5 nr1r nr1r sin sin CHAPTER TWO...REFRACTION AND REFLECTION Example 3: Light travels from air into an optical fiber with an index of refraction of (a) In which direction does the light bend? (b) If the angle of o incidence on the end of the fiber is 22P P, what is the angle of refraction inside the fiber? (c) Sketch the path of light as it changes media. Solution: (a) Since the light traveling from a rarer region (lower n) to a denser region (higher n), it will bend toward the normal. (b) We will identify air as medium 1 and the optical fiber as medium 2. Thus, 1, nr2 R 1.44 and the angle of incidence is θr1r 22. Snell's law then becomes: θr1r nr2r θr2 1 sin sin θr2 sin θr2 R 0.37 / θr2r -1 sinp P (0.26) 15 (c) The path of the light is shown in the figure below Example 4: A coin is at the bottom of a swimming pool of depth 2 m. The refractive index of air and water are 1 and 1.33, respectively. What is the apparent depth of the coin? Solution: nrar 1, nrwr 1.33 Dr. Anees Ali Page 5

6 Where: AB: apparent depth AC: actual depth From the diagram, Δ ABD Δ ACD tan r AAAA AAAA tan i AAAA AAAA By considering only small angles of r and i, hence n 2 then tan ii tan rr From the Snell's law tan r sin r and tan i sin i sin ii AAAA AAAA AAAA sin rr AAAA AAAA AAAA sin ii nn 2 sin rr nn 1 AAAA nn aa AAAA nn ww AB 1.5 m nn aa nn ww Note: (important) Other equation for absolute refractive index in term of depth is given by: rrrrrrrr dddddddd h n aaaaaaaaaaaaaaaa dddddddd h Example 5: (H.W) A beam of light traveling in air has an angle of incidence of 42º when it enters a medium which has an index of refraction of What is the angle of refraction of the beam inside the medium? Ans: 33 θ Example 6: (H.W) A light beam travels at ms -1 in quartz. The wavelength of the light in quartz is 355 nm. a. Find the refractive index of quartz. b. If the same light travels through air, what is its wavelength there? (Given the speed of light in vacuum, c ms -1 ). Ans: 1.55, 550 Dr. Anees Ali Page 6

7 2.2: Reflection Definition is defined as the return of all or part of a beam of particles or waves when it encounters the boundary between two media. Laws of reflection: - The reflected ray lies in the plane of incidence. - The angle of reflection equals the angle of incidence. i r There are two types of reflection: 1. Specular reflection: in which all the light travelling in one direction and reflected in one direction (as in mirror). 2. Diffuse reflection: is the reflection of light from a surface such that an incident ray is reflected at many angles rather than at just one angle as in the case of specular reflection. The visibility of objects, excluding light-emitting ones, is primarily caused by diffuse reflection of light: it is diffusely-scattered light that forms the image of the object in the observer's eye. U2.3: Total internal reflection When light crosses an interface into a medium with a higher index of refraction, the light will bends toward the normal. Conversely, light traveling across an interface from higher n to lower n will bend away from the normal. At some angle, known as the critical angle, light traveling from a medium with higher n to a medium with lower n will be refracted at 90 (refracted along the interface). If the light hits the interface at any angle larger Dr. Anees Ali Page 7

8 than this critical angle, it will not pass through to the second medium at all. Instead, it will be reflected back into the first medium, this is known as total internal reflection. The critical angle can be found from Snell's law: n 1 sin i n 2 sin r n 1 sin θ c n 2 sin 90 n 1 sin θ c n 2 sin θ c n 2 /n 1 θ c sin 1 nn 2 nn 1 Note: Because sine any angle cannot be greater than 1 so (n 2 / n 1 ) 1, therefore n 2 < n 1. Optical fibers are based on this principle of total internal reflection. An optical fiber is a flexible strand of glass. The light travels along the optical fiber, reflecting off the walls of the fiber. With a straight or smoothly bending fiber, the light will hit the wall at an angle higher than the critical angle and will all be reflected back into the fiber. 2.4: Fermat's principle Fermat's principle or the principle of least time is the principle that the path taken between two points by a ray of light is the path that can be traversed in the least time. This principle is sometimes taken as the definition of a ray of light. Note: Fermat's principle leads to Snell's law; when the sines of the angles in the different media are in the same proportion as the propagation velocities, the time to get from P to Q is minimized. Example 7 (H.W): Derive Snell's law from Fermat's principle. Dr. Anees Ali Page 8

9 2.5 Image formation by a plane mirror Figures 1 and 2 show the light ray radiating from a point object and vertical (extended) object then reflecting from reflected surface. (1) Point object (2) Vertical (extended) object Where: s: object distance sˈ: image distance y: object height yˈ: image height The properties of image formed by a plane mirror are: Virtual Upright or erect laterally reverse The object distance, s equals to the image distance sˈ Obey the law of reflection Same size where the liner magnification is given by IIIIIIIIII heeeeee htt,yyˈ M OOOOOOOOOOOO heeeeee htt,yy 1 Dr. Anees Ali Page 9

10 Example 8: Find the minimum vertical length of a plane mirror for an observer of 2 m height standing upright close to the mirror to see his whole reflection. How should this minimum length mirror be placed on the wall? Solution: By using the ray diagram as shown in figure blow AL 1 HHHH, AB 1 EEEE 2 2 The minimum vertical length of the mirror is given by h AL + LB h 1 HHHH + 1 EEEE 2 2 h 1 2 (HHHH + EEEE) h 1m Height of observer The mirror can be placed on the wall with the lower end of the mirror is halved of the distance between the eyes and feet of observer. Dr. Anees Ali Page 10

11 Example 9: A rose in a vase is placed 0.25 m in front of a plane mirror. A person looks into the mirror from 2 m in front of it. How far away from the person to the image of the rose? Solution: Object distance, s 0.25 m From the properties of the image formed by the plane mirror, s sˈ thus sˈ 0.25 m. Therefore, the distance between the person and the image of the rose is given by: xx 2 + sˈ xx 2.25 mm Dr. Anees Ali Page 11