Question

A thin film of glass of refractive index 1.5 is inserted in one arm of Michelson
Interferometer. Calculate the thickness of the film if a shift of 10 fringes is observed.
The wavelength of the light used is 589 nm.

Answer 1

Answer on Question #74821, Physics / Optics

Question. A thin film of glass of refractive index 1.5 is inserted in one arm of Michelson Interferometer. Calculate the thickness of the film if a shift of 10 fringes is observed. The wavelength of the light used is 589 nm.

Given. $n = 1.5$; $n_0 = 1$; $k = 10$; $\lambda = 589$ nm.

Find. $l - ?$

Solution.

For Michelson Interferometer

without the film

$$\Delta_1 = 2 d n_0 - 2 t n_0 = m_1 \lambda$$

with the film

$$\Delta_2 = 2 l n + 2 (d - l) n_0 - 2 t n_0 = m_2 \lambda$$

$d, t$ – Michelson Interferometer arms lengths.

So,

$$\Delta = \Delta_2 - \Delta_1 = 2 l n + 2 (d - l) n_0 - 2 t n_0 - 2 d n_0 + 2 t n_0 = (m_2 - m_1) \lambda$$ $$\Delta = 2 l n + 2 d n_0 - 2 l n_0 - 2 t n_0 - 2 d n_0 + 2 t n_0 = (m_2 - m_1) \lambda$$ $$\Delta = 2 l n - 2 l n_0 = k \lambda \rightarrow l = \frac{k \lambda}{2 (n - n_0)} = \frac{10 \cdot 589 \cdot 10^{-9}}{2 (1.5 - 1)} = 589 \cdot 10^{-8} \, m = 5.89 \, \mu m$$

Answer. $l = 5.89 \, \mu m$.

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