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PIT_PIT [208]
3 years ago
11

What is the minimum uncertainty in the vertical component of the momentum of each photon in the beam after the photon has passed

through the slit?
Physics
1 answer:
RoseWind [281]3 years ago
8 0
<h3><u>Minimum uncertainty in the vertical component of the momentum of each photon:</u></h3>

According to Heisenberg's Uncertainty principle, both the “position and velocity of the particle” cannot be measured exactly at the same time. The momentum of the particle equals the product of its mass and velocity. And it can be inferred that the “product of the uncertainties” in the “momentum and the position” of a particle equals \frac{h}{4 \pi}.  

Immediately after the photon has passed through the slit, given particle has a momentum uncertainty of \Delta P_{x} and its position uncertainty is \Delta x, then the minimum uncertainty in its momentum will be

\Delta P_{x} . \Delta x \geq \frac{h}{4 \pi}

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Explanation:

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d\ sin\theta=n\lambda

For third maxima,

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(b) For second dark fringe, n = 2

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Hence, this is the required solution.

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3 years ago
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