Optics Thin-film interference HW-158 A plate of glass (n = 1.5) is placed over a

Question

Optics Thin-film interference HW-158 A plate of glass (n = 1.5) is placed over a flat plate of plastic (n = 1.2). A thin film of water (n = 1.33) is trapped between the plates as illustrated in the diagram at right. 2, Glass (n= 1.5) Water film (n = 133) Plastic (n= 1.2) Light of frequency. 4.8 × 1014 Hz is incident from above at essentially normal incidence. The locations of the dark and bright fringes, as viewed from above, are indicated in the top view diagram at right. bottom surface of the plastic. Are the plates in contact at point A (where the plate appears dark) or at point B (where the plate appears bright)? Explain your reasoning. a. th { Determine the thickness of the gap between the plates at the end where the plates are not in contact. (Express your answer in mm.) Show all work. b.

Explanation / Answer

2. ng = 1.5

np = 1.2

nw = 1.33

f = 4.8*10^14 Hz

a. excluding reflections from top and bottom surface

light from glass to water does not undergo phase shift

light from water to plastic does not undergo phase shift

hence they constructively interfere where plates touch

hence

plates touch at B and not A

b. let the thickness at the end be t

now, from the figure this is n = 4th order dark fringe

lambda = c/f = 625 nm

path difference = nw*t

hence

for n = 4th order dark fringe

nw*t = (2n + 1)*lambda/2

t = 2.11466*10^-6 m

c. if the film pof water is replaced with air

light from glass to airdoes not undergoes phase shift

light from air to plastic undergoes phase shift

hecne there is a dark fringe where the plates meet

ii. in this case, nw becomes 1

so, for same thickness

n decreases

hence number of fringes decrease

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