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horrorfan [7]
3 years ago
12

A Michelson interferometer uses light with a wavelength of 452.446 nm. Mirror M2 is slowly moved a distance x, causing exactly 2

.2700×102 bright-dark-bright fringe shifts to be observed starting and ending with a central bright spot. If the mirror moved only half the distance, what would the final image be?
a. Bright spot.

b. Dark spot.

c. In between bright and dark spots.

d. Impossible to determine, more information needed.
Physics
1 answer:
I am Lyosha [343]3 years ago
3 0

Answer:

Option b. Dark spot

Explanation:

Michelson interferometer gives the path difference of the light as path difference, d as twice the distance moved or covered by the mirror M_{2}, x and is given as:

d= 2x

Since, its an even multiple, we obtain a bright fringe

now, when the move half the distance, i.e., \frac{x}{2}

Therefore, the path difference for half the distance \frac{x}{2} is:

d = x

As it is clear that its an odd multiple which correspond to dark spot as the final image

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Answer:

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2 years ago
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serious [3.7K]

Answer:

The answer to your question is:

a) t = 3.81 s

b) vf =  37.4 m/s

Explanation:

Data

height = 71.3 m = 234 feet

t = 0 m/s

vf = ?

vo = 0 m/s

Formula

h = vot + 1/2gt²

vf = vo + gt

Process

a)

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             71.3 = 0t + 1/2(9.81)t²

             2(71.3) = 9,81t²

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b)

      vf = 0 + (9.81)(3.81)

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