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Blababa [14]
4 years ago
6

Two beams of coherent light travel different paths, arriving at point P. If the maximum destructive interference is to occur at

point P, what should be the path difference between the two waves? Two beams of coherent light travel different paths, arriving at point P. If the maximum destructive interference is to occur at point P, what should be the path difference between the two waves? The path difference between the two waves should be one-half of a wavelength. The path difference between the two waves should be one wavelength. The path difference between the two waves should be four wavelengths. The path difference between the two waves should be one and one-quarter of a wavelengths. The path difference between the two waves should be two wavelengths. The path difference between the two waves should be one-quarter of a wavelength.
Physics
1 answer:
labwork [276]4 years ago
8 0

Answer:

Explanation:

Two beams of coherent light travel different paths, arriving at point P. If the maximum destructive interference is to occur at point P, what should be the path difference between the two waves?

The path difference between the two waves should be one and one-quarter of a wavelengths.

The path difference between the two waves should be two wavelengths.

The path difference between the two waves should be one-half of a wavelength.

The path difference between the two waves should be one wavelength.

The path difference between the two waves should be one-quarter of a wavelength.

The path difference between the two waves should be four wavelengths

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Four resistors (67 ohm, 83 ohm, 433 ohm, and 309 ohm in that order) are connected in series to a 7.92 V battery of negligible in
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Answer:

.737 v

Explanation:

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R = 67 + 83 + 433 + 309 = 892 ohm

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2 years ago
A positively charged wire with uniform charge density +λ lies along the x-axis and a negatively charged wire with uniform charge
Kisachek [45]

Answer:

\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]

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The electric field created by an infinitely long wire can be found by Gauss' Law.

\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E2\pi r h = \frac{\lambda h}{\epsilon_0}\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 r} \^r

For the electric field at point (x,y), the superposition of electric fields created by both lines should be calculated. The distance 'r' for the first wire is equal to 'y', and equal to 'x' for the second wire.

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3 years ago
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