2-Slit Intensity Pattern

1. Sep 29, 2010

xago

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

$$\psi$$2 = A*ei(kr1 - wt)

$$\psi$$1 = A*ei(kr2 - wt)

3. 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?

2. Sep 29, 2010

hikaru1221

This assumption still works well for double-slit interference experiment

You should check your calculation. Areduced = A/sqrt(100).