Calculating visibility of interference fringe

In summary, the homework statement is that 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}}
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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?
 
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  • #2
Should I post this somewhere else?
 
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  • #3
This work is intended as a help to show, using very simple geometrical models, how http://www.visibilityacceleration.com [Broken] is built into the Young's interference patterns produced by incoherent sources.
 
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1. How do you calculate visibility of interference fringe?

To calculate the visibility of interference fringe, you need to measure the maximum and minimum intensities of the fringe pattern. Then, use the formula V = (Imax - Imin) / (Imax + Imin) where V is the visibility and Imax and Imin are the maximum and minimum intensities, respectively.

2. What is the significance of calculating visibility of interference fringe?

Calculating the visibility of interference fringe is important as it provides information about the quality of the interference pattern. A high visibility value indicates a well-defined and clear interference pattern, while a low value may indicate factors such as poor experimental setup or external disturbances.

3. What factors can affect the visibility of interference fringe?

Several factors can affect the visibility of interference fringe, including wavelength of light, distance between the interfering sources, and the quality of the beamsplitter or mirrors used. External factors such as vibrations or temperature changes can also impact the visibility of the interference pattern.

4. How can you improve the visibility of interference fringe?

To improve the visibility of interference fringe, you can use a monochromatic light source with a narrow bandwidth, increase the distance between the interfering sources, and use high-quality optical components. It is also important to minimize external disturbances and ensure proper alignment of the experimental setup.

5. Can visibility of interference fringe be greater than 1?

No, visibility of interference fringe cannot be greater than 1. It is a dimensionless quantity that ranges from 0 to 1, with 1 representing a perfect interference pattern with no loss of intensity. A value greater than 1 would indicate an error in measurement or calculation.

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