What is the resulting intensity maximum at a minimum?

[xago]

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

$$\psi$$2 = A*e^{i(kr₁ - wt)}

$$\psi$$1 = A*e^{i(kr₂ - wt)}

The Attempt at a Solution

I'm assuming that we add the 2 wave equations to find $$\psi$$total.
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?

 

[hikaru1221]

I'm assuming that we add the 2 wave equations to find $$\psi$$total.

This assumption still works well for double-slit interference experiment

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. A_reduced = A/sqrt(100).