Question

In a Michelson interferometer by moving the mirror through a distance of x/4. The path difference is:​

Answer 1

Answer:for example

Answer

A shift of one fringe corresponds to a change in the optical path length of one wavelength. When the mirror moves a distance d, the path length changes by 2d since the light traverses the mirroe arm twice. Let N be the number of fringes shifted. Then, 2d=Nλ and

λ=

N

2d

=

792

2(0.233×10

−3

m)

=588×10

−7

m=588 nm

Answer 2

Answer:

Concept :

The best illustration of an amplitude-splitting interferometer is the Michelson interferometer. Coherent beams are created in the Michelson interferometer by splitting a light beam coming from a single source with a beam splitter, a partially reflecting mirror. Ordinary mirrors then redirect the resulting reflected and transmitted waves to a screen, where they superimpose to produce fringes. This is referred to as interference by amplitude division. This interferometer was used in 1817 in the well-known Michelson-Morley experiment, laying the door for the Special theory of Relativity by proving there was no electromagnetic wave-carrying ether.

Explanation:

• A transverse wave is light. The amplitudes of two waves that travel through the same medium at the same wavelength and amplitude mix.
• The end outcome will be a wave that is either larger or smaller than the original wave.
• Interference is the term for the increase in amplitudes caused by the superposition of two waves.
• The waves are said to interfere destructively when the crest of one wave collides with the trough of the other, producing a resultant intensity of zero.
• In contrast, if the crests of two waves collide, the consequence will be maximum intensity, and this is when the waves are said to interfere positively.
• If one of the mirrors is moved through a distance λ/4, the path difference changes by λ/2 and a maximum is obtained.

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