Question

When a thin film of a transparent material of refractive index 1.5 for wavelength 5890 Å is inserted in one of the arms of a Michelson's interferometer, a shift of 65 circular fringes is observed. Calculate the thickness of the film.

Answer 1

For Michelson interferometer without the film

$∆_1=2dn_0—2tn_0=m_1λ (1)$

with the film

$∆_2=2ln+2(d-l)n_0-2tn_0=m_2λ (2)$

where d and t is Michelson interferometer arms lengths

Using (1) and (2) we got

$∆=∆_2-∆_1=2ln-2ln_0=kλ (3)$

In our, case we have k=65, λ=5890×10^-10 m, n=1.5, n=1

l=38.2×10^-6 m

Answer

38.2×10^-6 m