This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 Halfwavelength 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 yellowgreen 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 onehalf 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 onehalf 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....
View
Full
Document
This note was uploaded on 04/18/2008 for the course ENGLISH 112 taught by Professor Loughlin during the Winter '08 term at UBC.
 Winter '08
 Loughlin

Click to edit the document details