Can a Single Point Grapefruit Create an Interference Pattern?

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SUMMARY

The discussion centers on the misconception that a single point grapefruit can create an interference pattern. The calculations involving the de Broglie wavelength (λ = h/p) and the small angle approximation (sin(α) = λ/d) were incorrectly applied, leading to confusion. It is established that a single particle, such as a grapefruit, cannot produce an interference pattern; rather, a multitude of particles is required to observe such a phenomenon. The conversation highlights the importance of understanding quantum mechanics principles and the nature of wave-particle duality.

PREREQUISITES
  • Understanding of de Broglie wavelength (λ = h/p)
  • Familiarity with small angle approximation in trigonometry
  • Knowledge of wave-particle duality in quantum mechanics
  • Basic principles of interference patterns in physics
NEXT STEPS
  • Research the concept of wave-particle duality in quantum mechanics
  • Learn about low-intensity interference experiments and their implications
  • Study the mathematical derivation of the de Broglie wavelength
  • Explore the principles of quantum measurement and wave function collapse
USEFUL FOR

Students of physics, educators teaching quantum mechanics, and anyone interested in the principles of wave-particle duality and interference patterns.

stuDYING
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Homework Statement
Even a grapefruit has a de Broglie wavelength.

If the grapefruit is thrown with a speed of 8 m/s toward a wall with two holes separated by 0.6 m, find the angular separation between successive maxima of the resulting interference pattern. Treat the grapefruit as point-like and assume its mass is 0.6 kg. Planck’s constant is 6.62607 × 10^−34 J · s. Answer in units of radian.
Relevant Equations
lambda = h/p
I tried using lamba = h/p as follows:

(6.626 * 10^-34 J *s) / (8 m.s * 0.6 kg) = 1.38041667*10^−34

and then using the small angle approximation sin(alpha) = lamba/d as follows:

(1.38041667*10^−34)/(0.6m) = 2.30069444 * 10^−34

then converting to radians with the following:

(2.30069444 * 10^−34) * (pi/180) = 4.01546931* 10^−36

However, this answer is none of the answer choices so I did something wrong. I think I might have not been supposed to use the small angle approximation but I'm not sure if that's the only thing and if so how to fix it.
 
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Hey there. I think you converted the ##sin(\alpha)## into radians, not the actual angle ##\alpha##. How do you change ##sin(x)## to ##x##.?
 
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osilmag said:
Hey there. I think you converted the ##sin(\alpha)## into radians, not the actual angle ##\alpha##. How do you change ##sin(x)## to ##x##.?

Oh gosh I'm so dumb sometimes. You're completely right - thank you so much!
 
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I don't mind preposterous problems, but I do mind nonsense problems that try to be cute but actually undermine what is to be reinforced. It's OK to suspend disbelief and go along with the gag of a "point" 0.6-kg grapefruit. However, the question asserts that there is a "resulting interference pattern." Huh? That's not OK. A single point grapefruit does not make an interference pattern; it makes a single (point) splat on the observation screen.

One needs to use many many point grapefruits to see the interference pattern eventually emerge on the screen, in this case as streaks of grapefruit juice. The gradual emergence of the interference pattern, one particle at a time, has been experimentally verified with low-intensity interference experiments. Whether you believe that a photon interferes with itself or not, when it hits that screen and you measure its position BAM! it collapses to a point in space where the probability of being there is high. This question promotes the wrong picture that a single particle generates a complete interference pattern and that one observes on the screen a superposition of identical patterns, each produced by a single particle. Tsk! Tsk! Tsk! :rolleyes:
 
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