Determination of Refractive Index or Thickness of Thin Transparent Plate or Film

A thin transparent film of thickness \(t\) and refractive index \(\mu\) is placed normally to the light ray between the mirror \(M_2\) and the beam splitter plate. Due to the insertion of the film, the optical path of the light ray increases by

\[ 2(\mu - 1)t \]

Consequently, the white fringe is displaced from its original position. The displacement of mirror \(M_1\) is adjusted with the help of a micrometer screw until the cross-wire of the telescope again coincides with the white fringe.

Let the displacement of mirror \(M_1\) be \(x\). The optical path difference produced due to the insertion of the thin film becomes equal to the optical path difference produced by the displacement of mirror \(M_1\).

Since the light travels twice between the beam splitter and the mirror, the path difference due to the displacement \(x\) is

\[ 2x \]

Therefore,

\[ 2(\mu - 1)t = 2x \]

Dividing both sides by 2, we get

\[ (\mu - 1)t = x \]

Hence,

\[ \boxed{t = \frac{x}{\mu - 1}} \]

Alternatively,

\[ \boxed{\mu - 1 = \frac{x}{t}} \]

Result:
If the thickness \(t\) of the thin film and the displacement \(x\) of the mirror are known, the refractive index \(\mu\) of the material can be determined using the above equation.