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Transcript

Phase Change on Reflection To understand interference caused by multiple reflections it is necessary to consider what happens when a light wave moving in air hits a material such as glass. The reflected pulse is said to undergo a phase change of 180° or π radians. The reflected pulse is 180° out of phase with the incident pulse. If these two pulses were to meet they would momentarily cancel as they passed one another. This happens whenever light waves are reflected from a material with a higher refractive index. If the waves are reflected from a lower refractive index material there is no phase change. The importance of phase change is that the optical path of the reflected ray is changed. A phase shift of π radians is equivalent to having travelled an extra λ/2 distance compared to a wave that had not been reflected. Thin (parallel) film interference When we look at a thin film of oil on water we see colours. This is caused by interference produced by reflection from the air/oil boundary and the oil/air boundary. 1 air oil water 2 n=1 d n=1.45 n=1.33 At each boundary some light is reflected and some refracted. This is called division by amplitude. Someone looking at rays 1 and 2 would see an interference pattern. This is caused by path difference between the rays. 1 2 air n=1 oil d n=1.45 If we assume the angle of incidence to be 0 degrees, then the extra distance travelled by ray 2 will be 2d. This means that the optical path difference will be equal to 2nd However, there will be a phase change n=1.33 of λ/2 at the first boundary (ray 1), since the ray is being reflected by a layer of greater refractive index water The total optical path difference will be equal to? = 2nd + λ/2 For constructive interference the path difference must be equal to a whole number multiple of wavelengths 2nd +λ/2 = mλ 2nd = mλ – λ/2 = (m + ½) λ 1 d (m )λ 2 2n This equation gives the thickness at which light of wavelength, λ will produce constructive interference. For destructive interference the path difference must be equal to an odd number of half wavelengths. 2nd + λ/2 = (m + ½ ) λ 2nd = mλ + λ/2 – λ/2 mλ d 2n For destructive interference, m=1 for minimum thickness. Example Calculate the minimum thickness of oil which will produce destructive interference in green light of wavelength 525nm. d mλ 2n 1 525 109 d 2 1.45 1.81 107 m