# Intensity of superimposed waves at the michelson inferometer

The Problem is about the Michelson Inferometer described by the picture below

Let the Beam Splitter be one, who reflects half of the intensity and transmits the other half of it.

Furthermore assume an initial wave with amplitude $E_0$ and intensity (mean value over time) $I_0$.

Then we can say that the intensity of the violet and the red wave just before the screen is $\frac{I_0}{4}$ respectively. Since the intensity is proportional to the amplitude squared, the amplitude of each wave is $\frac{E_0}{2}$. By regulating the mirrors accordingly, we can assume constructive interference of the waves and thus have an amplitude of $E_0$ and with it again the intensity of $I_0$. The effective area of the intensity didn't change and thus the energy at the screen is the same as at the beginning.

So far so good.

But when beam 1 hits the Beam Splitter for the second time half of it is reflected and similarly half of the beam 2 is transmitted by hitting the Beam Splitter a second time (not showed in this sketch). By going through the amplitudes, intensities and phase jumps we have again constructive interference of the two beams and the amplitude $E_0$ and intensity $I_0$. Again the effective area of the intensity hasn't changed with respect to the beginning and we have agin the same energy. Thus in total we doubled the energy of the system. And here, at the least, it seems strange to me...

At which point is my reasoning wrong?

• Think about the varying intensity of the fringes. And BTW there could be placed a second observation screen above the laser. The other half amount of the emitted photons will arrive on this screen. – HolgerFiedler Jul 22 '17 at 6:11

You must first work with amplitudes and not worry about the mean intensity.

Let the amplitude of the wave be $$A$$ as it leaves the laser.
If the intensity is reduced by a half on reflection or transmission at the beam splitter then the amplitude is $$\frac {A} {\sqrt 2}$$.

So after all the reflection two waves of amplitude $$\frac {A} {\sqrt 2}$$ end up on the screen in phase with one another.
Thus the total amplitude is $$\frac {A} {\sqrt 2} + \frac {A} {\sqrt 2}=\sqrt 2A$$ leading to an intensity which is twice that of the original laser but only at the centre of the screen.

If you look at this video at time 1:35 you will see that the demonstrator produces a large bright spot on the screen as a result of the constructive interference of the two beams.
At time 2:40 you will see that there is a difference between having only one beam shining on the screen and then having two beams the intensity being larger when the two beams interfere "constructively".

Note that the large spot is not uniformly illuminated and the intensity of light falls off as one moves away from the centre of the spot.
This is because the phase between the two beams when they arrive at the screen depends on the position at which they arrive on the screen.
This means that the average intensity across the screen is not twice the intensity of the laser light.

@HolgerFiedler has made a pertinent comment in that another pattern can be formed on another screen.

Continue watching the video from time 4:10 to see how the second patern second pattern can be produced, that pattern being the opposite to that produced on the screen in your diagram.
Note the contrast between the two patterns.
So where on one pattern is brighter the other pattern the other is darker.

Overall the Michelson interferometer does not produce extra light all it does is distribute the available light in different directions.

At the end of the video the demonstrator asks a question as to how it is that the two patterns are different.
To answer the question think about the difference between the reflection of light from an air to glass interface and from a glass to air interface.