Calculating visibility of interference fringe

[Bacat]

Homework Statement

A photomultiplier (PMT) is arranged to detect the photons from a double-slit experiment. It is placed at a point P in the detection plane and makes an angle \theta with the horizontal of one of the slits. Assume that the two slits have different widths and that the widths are much less than the wavelength of light, \lambda.

The probability amplitude for a single photon of wavelength \lambda to strike the PMT from one of the slits is \sqrt{2} more than for the other slit. Calculate the visibility of the interference fringes:

V=\frac{P_{max} - P_{min}}{P_{max}+P_{min}}

Where P_{max} is the maximum probability and P_{min} is the minimum probability that a photon is detected.

Homework Equations

A = A_1 + A_2 (amplitudes are summed)

P = (Abs(A_1 + A_2))^2 (probability is sum of amplitudes squared)

The Attempt at a Solution

I let A_1 be the greater amplitude, so it is equal to A_2 + \sqrt{2}. Then I use the equation for probability above to solve for A_2.

First problem:

I get different results if I solve it by hand or if I use Mathematica. Solving by hand I find that A_2 = A_1 = \frac{1-\sqrt{2}}{2}, but Mathematica tells me that A_2 = -1.20711 < 0 or A_2 = -0.207107 < 0

The answer Mathematica gives is in decimal format...but the fact that it comes up negative is really confusing me...I feel like the amplitude must be positive...?

Second problem:

If I take the answer I calculate by hand and add the amplitudes and square to find the probability I get:

P = A_1 + A_2 = (Abs(\frac{1-\sqrt{2}}{2} + \frac{1-\sqrt{2}}{2}))^2 = 3 - 2\sqrt{2} = 0.171573 < 1

The probability should sum to one, but it is much less than one.

What am I doing wrong?

 

[Bacat]

Should I post this somewhere else?

 

[Visibility]

This work is intended as a help to show, using very simple geometrical models, how http://www.visibilityacceleration.com is built into the Young's interference patterns produced by incoherent sources.