- #1

- 144

- 0

The total field at one output would be

[tex] \vert \vec{E}_1 + \vec{E}_2 \vert^2 = \vert E_1 \vert^2 + \vert E_2 \vert^2 + 2 Re\left( \vec{E}_1\cdot \vec{E}_2^{*} \right) [/tex]

The question is now, does it make sense to say that the interference only happens at the BS, and hence the interference term [itex] \vec{E}_1\cdot \vec{E}_2^{*} [/itex] is independent of the longitudinal distance after the BS, so that it doesn't matter where a detector is placed to detect the total beam?

Or will it be dependent on the longitudinal distance?

If that is the case (that it does depend), would it then mean that if you place a lens after the BS, the interference term would be changed, because the beam waist size is changed due to the lens?

But how can a lens change (the area integral of) the interference term (power from the interference), if it only transforms everything one-to-one to a different scale?