- #1

Salmone

- 101

- 13

With focus to the second interferogram, when the "optical path difference" (OPD from now on) is equal to ##0## of course we have a maximum of the intensity since both beams interfere constructively and when the movable mirror is moved so that the OPD is equal to ##\frac{\lambda_1}{2}## where ##\lambda_1## is the wavelenght of one of the beams, this one interfere desctructively with his part that traveled on the other path eliminating each other and the beam of wavelenght ##\lambda_2## also interfere with his part so that we get an intensity ##\neq 0## but smaller then the first one obtained when ##OPD=0##.

My question is:

At some points in the interferogram we see that total intensity of light is zero, in my opinion we get zero intensity when the movable mirror is placed at a point such that the OPD is equal to ##(n+\frac{1}{2})\lambda## for both beams, but this condition is only possible if ##\lambda_1=n\lambda_2## with ##n## being an integer,

**is this the only condition in which we get total intensity equal to zero?**

Are we always sure that there will be a certain position for the movable mirror such that the total intensity is equal to zero?

Are we always sure that there will be a certain position for the movable mirror such that the total intensity is equal to zero?