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The element sodium can emit light at two wavelengths, λ1=588.9950nm

and λ2=589.5924nm. Light from sodium is being used in a Michelson interferometer (Fig. 35-21). Through what distance must mirror M2 be moved if the shift in the fringe pattern for one wavelength is to be 1.00 fringe more than the shift in the fringe pattern for the other wavelength?

Short Answer

Expert verified

The distance moved by the mirror is 290.7μm.

Step by step solution

01

Given data

λ1=588.9950nmλ2=589.5924nm

02

Principal of Michelson interferometer

Fiber optic Michelson interferometer employs the same principle of splitting a laser beam and inserting the optical path difference between the arms.

03

Concept used

Existence of different wavelength shows the change in position of the mirror M2 because the path difference gives the distance at which the mirror is moved.

Express the relation for the distance at which the mirror M2is to be moved.

N1-N2=2L1λ1-1λ2

Here,N1 is the number wavelength in 2L thickness of air by the wavelength is N2, number of wavelengths in 2L thickness of air by the wavelength λ2 and L is the distance moved by, mirror M2.

Use the condition to express the distance moved by the mirrorM2.

N1-N2=1

04

Determine the distance moved by mirror

Rearrange the expression for distance moved by M2 in term of wave lengths.

L=121λ1-1λ2

Here L is the distance moved by mirror M2

Substitute 588.9950nm for λ1 and 589.5924nm for λ2 to find L.

L=121588.9950nm-1589.5924nm1nm10-9m=2.906×10-4m=290.7μm

Therefore, the distance moved by the mirror is 290.7μm.

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