Question

A Michelson interferometer uses light from a sodium lamp. Sodium atoms emit light having wavelengths 589.0 nm and 589.6 nm. The interferometer is initially set up with both arms of equal length (L1=L2), producing a bright spot at the center of the interference pattern. How far must mirror M2 be moved so that one wavelength has produced one more new maxima than the other wavelength?

Answer 1

The distance mirror M2 must be moved so that one wavelength has produced one more new maxima than the other wavelength is;

L = 57.88 mm

We are given;

Wavelength 1; λ₁ = 589 nm = 589 × 10⁻⁹ m

Wavelength 2; λ₂ = 589.6 nm = 589.6 × 10⁻⁹ m

We are told that L₁ = L₂. Thus, we will adopt L.

Formula for the number of bright fringe shift is;

m = 2L/λ

Thus;

For Wavelength 1;

m₁ = 2L/(589 × 10⁻⁹)

For wavelength 2;

m₂ = 2L/(589.6)

Now, we are told that one wavelength must have produced one more new maxima than the other wavelength. Thus;

m₁ - m₂ = 2

Plugging in the values of m₁ and m₂ gives;

(2L/589) - (2L/589.6) = 2

divide through by 2 to get;

L[(1/589) - (1/589.6)] = 1

L(1.728 × 10⁻⁶) = 1

L = 1/(1.728 × 10⁻⁶)

L = 578790.67 nm

L = 57.88 mm

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