# Intensity of interfering light waves

I had a conceptual question above light wave interference. Suppose that two light beams, each of an irradiance $I$ interfering on an area $A$ of a screen, such that all of the light from each beam falls on the same area $A$, and let the light beams be monochromatic, coherent, and have a zero phase difference. Then why is the resultant intensity there $4I$, and not $2I$?

Now i have read a similar question here which says that it should be but it doesn't help me understand the flaw in my reasoning which goes as follows. For energy conservation law to hold, we can only obtain a net power of $2I$ on the said spot, because $I_{net}=(P+P)/A$, where $P$ is the power emitted by each beam. If this is not true, then where does the extra power of $2P$ come from?

• You might find that the (many!) questions about energy and interference provide a different way to conceptualize this. – dmckee --- ex-moderator kitten Apr 9 '16 at 17:10
• @dmckee Ah! I was searching the wrong keywords then. But still, all those cases consider two waves that are split and then recombined, or cases where there is constructive interference at some places and destructive at others. What happens when the above case is considered (no beam splitters or possibility of destructive interference)? I don't understand how i can apply knowledge from those answers here! – FreezingFire Apr 9 '16 at 17:35
• @FreezingFire: you have essentially hypothesized the effects of a beam splitter/recombiner in your thought experiment. – Peter Diehr Apr 9 '16 at 19:10
• @PeterDiehr Well then where does the extra energy come from? In the beam splitter experiment I understand that the extra energy was the one which was not received by the other detector. There is no other detector here! – FreezingFire Apr 10 '16 at 2:50