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oee [108]
3 years ago
7

A technician places a fluid substance, with a refractive index of 1.61, between two horizontal panes of flat glass (each of whic

h has n = 1.45). She then illuminates the assembly from above with monochromatic light (λ = 330 nm). What is the minimum thickness (in nm) that the liquid layer must have, if the light is to be strongly reflected back toward its source?
Physics
1 answer:
Ahat [919]3 years ago
5 0

Answer:

the minimum thickness (in nm)  t = 102.5 nm

Explanation:

t = thickness of the film

n = index of refraction = 1.61

m = order = 1

∝ = wavelength = 330 x 10-9 m

so we have the formula

2nt = m∝

2 (1.61) t = (1) (330 x 10-9)

then...

t = 1.025 x 10-7 m

converting...

t = 102.5 nm

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Answer:

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Explanation:

From the question we are told that

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Let assume that the initial value of the emitted frequency is

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Hence new frequency will be  f_n  =  1 +  0.016 = 1.016

Generally from Doppler shift equation we have that

         f_1 =  [\frac{ v \pm v_o}{v \pm + v_s } ] f

Here v  is the speed of sound with value  v =  343 \ m/s

         v_s is the velocity of the sound source which is v_s = 0 \ m/s because it started from rest

         v_o  is the observer velocity So

Generally given that the observer id moving towards the source, the Doppler frequency becomes

                   f_1 =  [\frac{ v + v_o}{v + 0 } ] f

=>                1.016  =  [\frac{ 343  + v_o}{343 } ] * 1  

=>                v_o =  5.488 \  m/s

         

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