Estimating the Probability of Two Photons Landing on the Same Atom in a Double-Slit Experiment
You're asking how small the probability is that two photons, arriving one at a time,
land on the exact same atomic-scale location in a double-slit experiment.
To estimate this, we need to consider:
- A typical geometry for the double-slit setup (slit separation, distance to the screen).
- An estimate of the interference pattern area on the screen.
- The size of an atom as a target area.
- The probability density function of photon impacts based on quantum mechanics.
1. Estimate the Wavefunction Spread on the Screen
In a classic double-slit experiment, the interference pattern on the screen depends on:
- Slit separation: d
- Wavelength of light: λ
- Distance from the slits to the screen: L
The angle θ to the first bright fringe is approximately:
\[
\theta \approx \frac{\lambda}{d}.
\]
At a distance L from the slits, the transverse displacement y of that fringe is:
\[
y \approx L \cdot \theta = L \,\frac{\lambda}{d}.
\]
Example:
- \( d = 10^{-4} \,\text{m} \) (0.1 mm)
- \( \lambda = 5 \times 10^{-7} \,\text{m} \) (500 nm)
- \( L = 1 \,\text{m} \)
Plugging in these values:
\[
y \approx 1\,\text{m} \times \frac{5 \times 10^{-7}\,\text{m}}{10^{-4}\,\text{m}}
= 5 \times 10^{-3}\,\text{m} = 5\,\text{mm}.
\]
The central bright region might span about twice this distance (±5 mm), giving a total width of about
1 cm. Assuming a similar vertical extent, the area of the main interference region is:
\[
A_{\text{pattern}} \approx (1\,\text{cm}) \times (1\,\text{cm})
= 1\,\text{cm}^2
= 10^{-4}\,\text{m}^2.
\]
2. Compare This to an Atom-Sized Target
- An atom has a diameter of about \( 10^{-10}\,\text{m} \) (0.1 nm).
- Its approximate cross-sectional area is \( \sigma \approx 10^{-20}\,\text{m}^2 \).
3. Probability of One Photon Landing on an Atomic Spot
If the interference pattern is roughly uniform across
\( A_{\text{pattern}} = 10^{-4}\,\text{m}^2 \), then the probability of
one photon hitting that single atomic cross-section
\( \sigma \approx 10^{-20}\,\text{m}^2 \) is:
\[
P_{\text{one photon}}
= \frac{\sigma}{A_{\text{pattern}}}
= \frac{10^{-20}\,\text{m}^2}{10^{-4}\,\text{m}^2}
= 10^{-16}.
\]
Thus, for any given photon, the chance of hitting one atom-sized spot is
about \( 10^{-16} \) (one in \( 10^{16} \)).
4. Probability of Two Independent Photons Landing on the Same Atom
Assuming photons arrive one at a time and are independent, the probability of
two photons hitting the same atom is:
\[
P_{\text{two photons}}
= \bigl(P_{\text{one photon}}\bigr)^{2}
= (10^{-16})^2
= 10^{-32}.
\]
That's a one in \( 10^{32} \) chance—extremely small.
Conclusion
-
The probability of two photons landing on the exact same atom in the main interference
region is about \( 10^{-32} \)—effectively zero for most practical purposes.
-
Even with trillions of photons, the likelihood of two hitting the same atomic spot
remains vanishingly small.
-
For larger targets (like a grain in photographic film), the probability would be higher
but still relatively small.
🚀 Would you like a numerical simulation to visualize this probability in an actual setup?
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