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## 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 [tex]\theta[/tex] 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, [tex]\lambda[/tex].

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

[tex]V=\frac{P_{max} - P_{min}}{P_{max}+P_{min}}[/tex]

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

## Homework Equations

[tex]A = A_1 + A_2[/tex] (amplitudes are summed)

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

## The Attempt at a Solution

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

**First problem:**

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

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:

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

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

What am I doing wrong?