Controlling output light power by constructive/destructive interference

  • #1
Mayan Fung
131
14
In optical communications, one of the modulation methods is to control the optical power (Simplest case, for example, bright = bit 1, dim = bit 0). I learned that we can achieve this by a Mach-Zehnder modulator (MZ modulator).

Simply speaking, the principle of MZ modulator is to split the input light into two, traveling in two waveguides. By applying a voltage across one of the two arms, we can alter the refractive index of it and thus inducing a phase difference between the two split lights. Then, when they are combined as one output light signal, if they constructively interfere, then the output power is maximum (bit 1). Otherwise, if they destructively interfere, then the output power is minimum (bit 0).

It sounds good till here. Yet, I am puzzled about energy conservation. Let's say we split the input light into two with the same amplitude. When they combined with destructive interference, there is no light. How come we have optical power input but no power output?

1596513844585.png
 

Answers and Replies

  • #2
Baluncore
Science Advisor
12,081
6,201
The two point sources are combined so as to get cancellation in one direction.
There will be other directions in which the energy is summed.
 
  • #3
Mayan Fung
131
14
The two point sources are combined so as to get cancellation in one direction.
There will be other directions in which the energy is summed.

But in the above schematic, seems that there is only one output direction. Can the light escape the wavelength?
 
  • #4
Baluncore
Science Advisor
12,081
6,201
But in the above schematic, seems that there is only one output direction.
The diagram is symbolic.
It is not shown how light is split into two channels.
It is not shown how light is recombined into one line.
 
  • Like
Likes sophiecentaur
  • #5
sophiecentaur
Science Advisor
Gold Member
27,847
6,339
The diagram is symbolic.
It is not shown how light is split into two channels.
It is not shown how light is recombined into one line.
One possibility is to split and combine with half silvered mirrors. The diagram of a beam splitter is on this Wiki page. There are two input ports and two output ports. When combining your two beams, one port will let through the sum and the other will let through the difference resultants. You 'just' need to get the path lengths right to achieve that.
 
  • #6
Mayan Fung
131
14
I see your points. That makes sense if there are two outputs. Thanks
 

Suggested for: Controlling output light power by constructive/destructive interference

Replies
5
Views
398
Replies
54
Views
2K
Replies
3
Views
11K
Replies
8
Views
603
Replies
1
Views
212
  • Last Post
Replies
22
Views
571
Replies
1
Views
373
Replies
4
Views
2K
Top