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© John Parkinson 1
© John Parkinson 2 I think we are being watched! Refraction is the Bending of Waves due to a Change in Velocity Refraction is the Bending of Waves due to a Change in Velocity I wasn’t going to jump anyway
© John Parkinson 3 air glass / water REFRACTION
© John Parkinson 4 Optically less dense medium (1) Optically denser medium (2) Waves travel SLOWER Wavelength REDUCED Frequency UNCHANGED REFRACTIVE INDEX FROM 1 TO 2
© John Parkinson 5 1 2 normal Θ 1 = angle of incidence Refracted ray Θ 2 = angle of refraction LAWS OF REFRACTION 1. The incident ray, the refracted ray and the normal all lie in the same plane. 2. For two given media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant. SNELL’s LAW Incident ray
© John Parkinson 6 Incident ray normal Refracted ray 1 2 SNELL’s LAW
© John Parkinson 7 1 2 Absolute refractive indices are measured with respect to the velocity of light in a vacuum, which is very nearly the same as the velocity of light in air. [ n air = 1.0008 ] What will be the refractive index for rays of light travelling from medium 1 to medium 2, when 1 is not air? HENCE SO
© John Parkinson 8 water glass θ 30 0 n water = 1.33 c = 3.0 x 10 8 m s -1 c glass = 2.0 x 10 8 m s -1 Find angle θ QUESTION
© John Parkinson 9 Violet light has a higher refractive index in glass than red light
© John Parkinson 10
© John Parkinson 11 When light travels from an optically denser medium to a less dense medium, rays are bent away from the normal. A little light is also reflected. For a larger angle of incidence, more light is reflected as well as the refracted ray being bent further from the normal c For angles of incidence greater than the critical angle, TOTAL INTERNAL REFLECTION occurs. When the critical angle, c, is reached it produces an angle of refraction of 90 0 θθ N.B. The angle of incidence = The angle of REFLECTION
© John Parkinson 12 Critical angle depends upon the refractive indices of the media By Snell’s Law HENCE 1 2 C If medium 2 is air, n 2 = 1, and so
© John Parkinson 13 1 2 C AIR WATER For Water C = 48.8 0 For Crown Glass C = 41.8 0
© John Parkinson 14 observer PERISCOPE – two right isosceles prisms 45 0 Critical Angle for common glass = 42 0 Total internal reflection
© John Parkinson 15 Objective lens Eyepiece lens PRISMATIC BINOCULARS
© John Parkinson 16 THE MIRAGE Hot Desert Sand air layers Air layers closer to the sand are hotter and less dense. Light from the sky is successively bent until a critical angle is reached and then total internal reflection occurs. A mirage "water" illusion is seen because the mind initially interprets the light rays reaching our eyes as having come along a straight path originating from the ground. Thus, the image of that patch of sky we see "on the ground" is interpreted as a surface "pool of water." Water for my hump!
© John Parkinson 17 FIBRE OPTICS Light trapped in fibre core STEP INDEX FIBRE Core Cladding 1.47 1.50 A critical angle will exist for rays travelling from the core to the cladding core cladding Diameter around 50 μm Step Index fibre
© John Parkinson 18 core n 1 cladding n 2 c Suppose n1 n1 = 1.50 and n2 n2 = 1.47, then The light entering a fibre end must be inside a 'cone of viewing' if it is to be transmitted along the fibre, otherwise it passes through the core-cladding boundary because its angle of incidence at this boundary is greater than the critical angle. The acceptance angle, θ, of the cone of viewing, is given by the equation sin θ = (n 1 2 – n22 n22 ) 1/2 / nono where no no is the refractive index of the substance outside the fibre. If the acceptance angle is too small, the cone of viewing is too narrow. θ
© John Parkinson 19 A COHERENT BUNDLE: A bundle of optical fibres in which the relative spatial coordinates of each fibres are the same at the two ends of the bundle. Such a bundle are used for the transmission of images. A NON-COHERENT FIBRE bundle, as you would expect, does not have this precise matrix alignment since they need only transmit light for illumination purposes. They are cheaper to produce.
© John Parkinson 20 End probe containing coherent bundle, incoherent bundle, lens and surgical instruments Controls Eyepiece Light injected here ENDOSCOPE This is an endoscope image of the inside of the throat. The arrows point to the vocal chords
© John Parkinson 21 FIBRE OPTIC COMMUNICATIONS core cladding PROBLEM: Input light rays cannot be precisely parallel. Rays taking different paths will take different times to travel along the fibre, resulting in the jumbling of the signal. Solution : Monomode fibre only 5μm in diameter
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