Application of Michelson’s Interferometer: Measurement of the Difference Between Two Spectral Lines

Michelson’s interferometer can be used to determine the difference in wavelengths (or frequencies) between two closely spaced spectral lines. This method is based on the periodic change in visibility of interference fringes produced by the two wavelengths.

Let the wavelengths of two spectral lines be:

\[ \lambda_1 \quad \text{and} \quad \lambda_2 \]

where \( \lambda_1 \) and \( \lambda_2 \) are very close to each other.

Principle

When light of two nearly equal wavelengths is used in a Michelson interferometer, two sets of circular interference fringes are formed. Due to the slight difference in wavelengths, the bright fringes of one system gradually coincide with the dark fringes of the other system. Consequently, the fringes alternately become:

• Sharp and distinct (maximum visibility)
• Faint or invisible (minimum visibility)

The difference in wavelength can be calculated by measuring the displacement of the movable mirror between two successive positions of maximum or minimum visibility.

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Derivation

Let the optical path difference between the two interfering beams be:

\[ \Delta = 2d \]

where \(d\) is the displacement of the movable mirror.

For a wavelength \( \lambda_1 \), the condition for a bright fringe is:

\[ 2d = m\lambda_1 \]

Similarly, for the second wavelength \( \lambda_2 \), the condition for a bright fringe is:

\[ 2d = n\lambda_2 \]

At a position where the two systems of fringes coincide:

\[ m\lambda_1 = n\lambda_2 \]

Now, after the mirror is displaced by a distance \(x\), the optical path difference changes by:

\[ 2x \]

For a wave of wavelength \(\lambda\), a path difference \(\Delta\) produces a phase difference:

\[ \phi = \frac{2\pi}{\lambda}\Delta \]

Therefore, for the two wavelengths, the phase changes are:

\[ \phi_1 = \frac{2\pi(2x)}{\lambda_1} \]

\[ \phi_2 = \frac{2\pi(2x)}{\lambda_2} \]

The relative phase difference between the two waves is therefore:

\[ \Delta\phi = \phi_1 - \phi_2 \]

\[ \Delta\phi = 2\pi(2x) \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) \]

The two interference patterns become maximum again when their relative phase difference changes by one complete cycle:

\[ \Delta\phi = 2\pi \]

Hence:

\[ 2\pi(2x) \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) = 2\pi \]

Cancelling \(2\pi\) from both sides:

\[ \boxed{ 2x \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) = 1 } \]

Simple Physical Meaning

The two wavelengths have slightly different frequencies or wave numbers. Therefore, as the optical path difference increases, their interference patterns slowly move out of phase.

• When their relative phase is \(\left(0, 2\pi, 4\pi, \ldots\right)\), the two fringe systems reinforce each other → maximum visibility.
• When their relative phase is \(\left(\pi, 3\pi, \ldots\right)\), one system is bright while the other is dark → minimum visibility.