Physics Assignment

Physics Assignment - Phys 122 Introductory Physics II...

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Unformatted text preview: Phys 122 Introductory Physics II TUTORIAL 11 Wave Optics KEY Thin Films. 1. Light of wavelength 691 nm (in vacuum) is incident perpendicularly on a soap film ( n = 1.33) suspended in air. What are the two smallest nonzero film thicknesses (in nm) for which the reflected light undergoes constructive interference? 1 film film film film 2 Extra distance Condition for Half-wavelength traveled by wave constructive interference net phase change in the film due to reflection 2 , 2 , 3 , t + = K 123 1 4 4 4 442 4 4 3 142 43 4 4 4 The smallest film thickness occurs when the condition for constructive interference is film . Then, the relation above becomes 1 1 film film film 2 4 2 = or t t + = Since vacuum film film n = (Equation 27.3), we have that 2 vacuum 1 1 1 film 4 4 4 film 691 nm 1.30 10 nm 1.33 t n = = = = The next smallest film thickness occurs when the condition for constructive interference is 2 film . Then, we have 2 vacuum 3 3 3 1 film film film 2 4 4 4 film 691 nm 2 = 2 or 3.90 10 nm 1.33 t t n + = = = = z 2. A nonreflective coating of magnesium fluoride ( n = 1.38) covers the glass ( n = 1.52) of a camera lens. Assuming that the coating prevents reflection of yellow-green light (wavelength in vacuum = 565 nm), determine the minimum nonzero thickness that the coating can have. 1 SOLUTION Since the coating is intended to be nonreflective, its thickness must be chosen so that destructive interference occurs between waves 1 and 2 in the drawing. For destructive interference, the combined phase difference between the two waves must be an odd integer number of half wavelengths. The phase change for wave 1 is equivalent to one-half of a wavelength, since this light travels from a smaller refractive index ( n air = 1 00 . ) toward a larger refractive index ( n film = 1 38 . ). 2 1 camera lens coating (film) air Similarly, there is a phase change when wave 2 reflects from the right surface of the film, since this light also travels from a smaller refractive index ( n film = 1 38 . ) toward a larger one ( n lens = 1 52 . ). Therefore, a phase change of one-half wavelength occurs at both boundaries, so the net phase change between waves 1 and 2 due to reflection is zero. Since wave 2 travels back and forth through the film and, and since the light is assumed to be at nearly normal incidence, the extra distance traveled by wave 2 compared to wave 1 is twice the film thickness, or 2 t . Thus, in this case, the minimum condition for destructive interference is 2 1 2 t = film The wavelength of light in the coating is film vacuum 565 nm 1.38 409 nm = = = n (27.3) Solving the above expression for t , we find that the minimum thickness that the coating can have is t = = = 1 4 1 4 film (409 nm) 102 nm ______________________________________________________________________________ 3. A mixture of yellow light (wavelength = 580 nm in vacuum) and violet light (wavelength = 410 nm in vacuum) falls perpendicularly on a film of gasoline that is floating on a puddle of water. For both wavelengths, the refractive index of perpendicularly on a film of gasoline that is floating on a puddle of water....
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This note was uploaded on 04/18/2008 for the course ENGLISH 112 taught by Professor Loughlin during the Winter '08 term at UBC.

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Physics Assignment - Phys 122 Introductory Physics II...

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