α=10′′≈4.85∗10−5rad
n=1.4 refractive index
b=0.5cm=5∗10−3m the distance between the successive fringes
OPD−optical path difference
λ wavelength of light
Let d the thickness of the film at the place of incidence of the beam
For a beam reflected from the upper surface:
OPD1=−2λ;
For a beam reflected from the inner surface of the film:
OPD2=2dn
OPD=2dn−2λ
Minimum interference condition:
OPD=kλ−2λ; k∈Z
2dn=kλ
the transition to the neighboring strip corresponds to a change in k per unit
2(d+d′)n=(k+1)λ
d′=2nλ
tanα=bd′
λ=2nbtanα
tanα≈α for small angles in radians
λ=2∗1.4∗5∗10−3∗4.85∗10−5=6.79∗10−7m
Answer: λ=6.79∗10−7m
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