Problem 1 A flat thin film of plastic (up 1.60) of uniform thickness covers a pi

Question

Problem 1 A flat thin film of plastic (up 1.60) of uniform thickness covers a piece of flat flint glass (n 1.52) Light with a variable wavelength ° is incident upon the plastic as shown in the incident light, air (n = 1.00) stic (np = 1.60) flint glass (n 1.52) [10 pts] a) When viewed from above, reflected light is a maximum (bright) when the wavelength ° of the incident light is 6.95 × 10.7 m. What is the smallest possible thickness of the plastic layer? what wavelength will the first minimum (dark) in reflected light occur? part (b), at what wavelength will the next maximum (bright) in reflected light [5 pts] b) As the wavelength of the incident light is decreased below 6.95 x 10' m, at (S ptsc) As the wavelength of the incident light is further decreased below that found in occur? [S pts) d) If the thickness of the plastic found in part (a) is tripled, for incident light with wavelength 6.95 x 10 m, will the reflected light be maximum (bright) or minimum (dark)?

Explanation / Answer

given

np = 1.6

ng = 1.52

wavelength == lambda

a. lambda = 6.95*10^-7 m

reflected light shows constructive interference

now, in the air plastic interface the light undergoes phase shift of pi/2 but not on the other interface

hence minimum thickness = t

2t*np = lambda/2

t = 108.59375 nm

b. for destructive interferece below 695 nm

2*t*np = n*lambda

lambda = 347.5 nm for n = 1

c. for next maxima

2*t*np = 3lambda/2

lambda = 231.66 nm

d. if t is tripled

t' = 3t

then for lambda = 695 nm

6t*np = lambda*n = 3*lambda/2

n = 3/2

hence there will be a constructive interferecne ( bright fringes)

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