A Visual Explanation of the Michelson InterferometerTara L.

This is a diagram of a generic Michelson interferometer

A laser beam (the source) shines through a lens into a partially-silvered mirror, called a beam splitter.

beam splitter

source

The beam splitter is just transparent enough to let half the light pass through to the fixed mirror

beam splitter

source

While still being silvered enough to reflect the other half of the light to the movable mirror

beam splitter

source

The two beams split by the beam splitter is reflected off of their respective mirrors, back through the beam splitter

beam splitter

source

And onto the detector

source

There is a compensator in between the fixed mirror and the beam splitter to make sure that when the mirrors are at equal distances from each other,

sourcethe two beams have equal optical path lengths.

compensator

Initially, the two mirrors are at equal distances from the beam splitter (d1=d2). There is no difference in path length, the light waves are in phase,

source

and the detector shows an image that is bright in the center

d1

d2

Image at detector

http://www.lightandmatter.com/html_books/0sn/ch06/figs/hw-sine-wave.png

As you move the movable mirror away, d2 will increase. When the path difference becomes ½ λ, the two waves undergo destructive interference,

and the image at the detector will be dark

d1d1

d2 d2

Image at detector

Image at detector

http://www.lightandmatter.com/html_books/0sn/ch06/figs/hw-sine-wave.png http://www.lightandmatter.com/html_books/0sn/ch06/figs/hw-sine-wave.png

Notice that because the light wave has to travel from the beam splitter to the movable mirror

And back again

The movable mirror only has to be adjusted by ¼ λ to increase the path travelled by the beam by ½ λ d2

d1

¼ λ

¼ λ

Δd = ¼ λ + ¼ λ

= ½ λ

Image at detector

http://www.lightandmatter.com/html_books/0sn/ch06/figs/hw-sine-wave.png

Since the waves differ by λ, they are once again in phase and result in constructive interference (and brightness in the image).

d2

d1

½ λ

½ λ

Δd = ½ λ + ½ λ

= λ

Image at detector

http://www.lightandmatter.com/html_books/0sn/ch06/figs/hw-sine-wave.png

Likewise, the movable mirror only has to be adjusted by ½ λ to change the path travelled by λ.

Here’s a quick question to make sure you’re getting this:

› Assume your laser has a wavelength of 560nm, and your mirrors start at equal distances from the beam splitter

› You shift your movable mirror by 1.5876mm

› What’s the path difference between your two mirrors now?

› Is the center of the image on your detector dark, or bright?

Solution: › Convert mm to nm: 1.5876mm = 1,587,600nm

› The light beam must travel to the movable mirror and back, so multiply the distance that the mirror shifted by 2: 1,587,600nm*2 = 3,175,200nm

› The path difference is 3,175,200nm

› Is this number an odd integer multiple of ½ λ (in which case it would cause destructive interference and result in darkness in the center),or an integer multiple of λ (which would cause constructive interference and a result in a bright center)?: 3,175,200nm/560nm = 5,670

› 5,670 is an integer, so it’s constructive interference resulting in a bright center on the image at the detector.