# Why is there destructive interference in one path in Wheeler's delayed choice experiment?

In the diagram below I've drawn Wheeler's delayed choice experiment, in the case where a 2nd half-silvered mirror has been inserted near the detectors.

---->\-------->\
|         |
v         v
\-------->\--> D2 (0%)
|
v
D1 (100%)


Every summary of this experiment I've found says that in this case destructive interference will occur at D2 and constructive interference will occur at D1.

I understand why the two paths interfere constructively at D1:

• One path has 1 reflection => phase change of 180$$^\circ$$
• The other path has 3 reflections => phase change of 3 x 180$$^\circ$$ which equals 180$$^\circ$$ modulo 360$$^\circ$$

I don't understand why the two paths interfere destructively at D2:

• One path has 2 reflections
• The other path also has 2 reflections

Can anyone explain why interference is supposed to be destructive at D2?

• I believe you drew the diagram wrong. D1 is getting some paths with one reflection and some with 3, so D1 should have the destructive interference and be 0%
– TKoL
Feb 10, 2021 at 17:19
• Yes, given the answer you appear to be right. I copied the diagram and the intensity split from quantamagazine.org/… Feb 11, 2021 at 9:21