Question

Monochromatic light of variable wavelength is incident normally on a thin sheet of plastic film in air. The reflected light is a maximum only for 477.1 nm and 668.0 nm in the visible spectrum.What is the minimum thickness of the film (n=1.58)?

Answer 1

Answer:

thickness t = 528.433 nm

Explanation:

given data

wavelength λ1 = 477.1 nm

wavelength λ2 = 668.0 nm

n = 1.58

solution

we know for constructive interference condition will be

2 × t × μ = (m1+0.5) × λ1     ....................1

2 × t × μ = (m2+0.5) × λ2     ....................2

so we can say from equation 1 and 2

(m1+0.5) × λ1 = (m2+0.5) × λ2

so

$\frac{\lambda 2}{\lambda 1} = \frac{m1+0.5}{m2+0.5}$     ..............3

put here value and we get  

$\frac{668.0}{477.1} = \frac{m1+0.5}{m2+0.5}$  

$\frac{m1+0.5}{m2+0.5}$   = 1.4

$\frac{m1+0.5}{m2+0.5} = \frac{7}{5}$   ...................4

so we here from equation 4

m1+0.5  = 7

m1 = 3    .................5

m2+0.5 = 4

m2 = 2    .................6

so now put value in equation  1

2 × t × μ = (m1+0.5) × λ1

2 × t × 1.58 = (3+0.5) ×  477.1

solve it we get

thickness t = 528.433 nm