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sattari [20]
3 years ago
9

You're using a monochromatic beam of light with wavelength 500 nm in an interferometer. What is the miminum distance you would n

eed to move one of the interferometer's mirrors to change a spot of constructuve interference to a spot of destructive interference?
Physics
1 answer:
Marina CMI [18]3 years ago
5 0

Answer:

Minimum distance needed to move one of the mirrors = 125 nm

Explanation:

Constructive interference occurs when the maxima (trough or crest) of two waves that are in phase add together so that the amplitude of the resulting wave is equal to the sum of the individual amplitudes.

Destructive interference occurs when the maxima (trough or crest) of two waves that are out of phase by 180⁰ or half a wavelength combine to produce a smaller amplitude than the individual amplitudes that combine. It can even produce a wave of zero amplitude.

Since the spot of constructive interference has been changed to destructive interference, Path difference = Wavelength/2

Path difference = 500nm/2 = 250 nm

Path difference = 250 * 10⁻⁹ m

Since there are two interferometer's mirrors and only one is moved, the minimum distance will be a half of the path difference.

That is, minimum distance = (path difference)/2

Minimum distance = (250 * 10⁻⁹)/2

Minimum distance = 125 * 10⁻⁹ m

Minimum distance = 125 nm

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From the question we are told that:

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A double-slit experiment is set up using red light (λ = 706 nm). A first order bright fringe is seen at a given location on a sc
Elanso [62]

Answer:

λ = 470.66 nm

Explanation:

for bright fringey_m = \frac{m\lambda D}{d}

D= distance between slit and screen

d= distance between the slits

for first order bright fringe m = 1,

        y_1 = \frac{1\lambda D}{d}

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for dark fringe,we have

y_m = {(m + 1/2)\lambda D}{d}    

Now to get the dark fringes at the same location we should have;

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6 0
4 years ago
A convex mirror, like the passenger-side rearview mirror on a car, has a focal length of -3.0 m . An object is 6.0 m from the mi
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From the ray diagram, we see that q has a distance of 2.0 m from the mirror, and it's on the other side of the mirror compared to the object, so

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This can also be verified by using the mirror equation:

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Therefore, this means that the image is virtual.

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