What is the resulting intensity maximum at a minimum?

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Homework Statement



Surprisingly, large interference effects can occur even when one of the interfering sources is
not very probable. In the two-slit interference experiment, if one slit is “stopped down” so that
the intensity of the wave getting through is reduced by a factor of 100 (relative to the other
slit), show that the intensity maximum of the pattern is still (roughly) 50 per cent higher than
at a minimum.

Homework Equations



\psi2 = A*ei(kr1 - wt)

\psi1 = A*ei(kr2 - wt)

The Attempt at a Solution



I'm assuming that we add the 2 wave equations to find \psitotal.
Since 1 slit has 1/100 the intensity of the other and since I is proportional to A2
then the amplitude of reduced-intensity wave would have A= 1/10000?
 
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xago said:

I'm assuming that we add the 2 wave equations to find \psitotal.

This assumption still works well for double-slit interference experiment :smile:

Since 1 slit has 1/100 the intensity of the other and since I is proportional to A2
then the amplitude of reduced-intensity wave would have A= 1/10000?
You should check your calculation. Areduced = A/sqrt(100).
 

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