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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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a bullet fired straight through a board 0.10 mm thick strikes the board with a speed of 480 m/sm/s, has constant acceleration th
vekshin1

the average acceleration of the bullet through the board   -55657×10⁵ m/s²

acceleration-  Rate of change of velocity with respect to time.

S = 0.10 mm = 10⁻⁵ m  ( distance)          [∵ 1mm = 10⁻³]

u =  480 m/s (initial velocity)

v = 345 m/s (final velocity)

As we know the 3rd equation of motion.  

v² = u² + 2aS

a = ? ( acceleration)

using these values in equation we get

(345)² = (480)² + 2×a× 10⁻⁵

a = (345)² -  (480)² / 2× 10⁻⁵

a =  -55657×10⁵ m/s²

The negative sign shows that the direction of acceleration is opposite of velocity thus bullet is slowing down.

the average acceleration of the bullet through the board  -55657×10⁵ m/s²

The given question is incomplete. The complete question is.

a bullet fired straight through a board 0.10 mm thick strikes the board with a speed of 480 m/sm/s, has constant acceleration through the board, and emerges with a speed of 345 m/sm/s.

What is the average acceleration of the bullet through the board/?

to know more about  the equation of motion :

brainly.com/question/13269040

#SPJ4

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2 years ago
An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms–1
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Answer:

Initial velocity of the object, u = 5 m/s

Final velocity of the object, v = 8 m/s

Mass of the object, m = 100 kg

Time take by the object to accelerate, t = 6 s

Initial momentum = mu = 100 — 5 = 500 kg m sˆ’1

Final momentum = mv = 100 — 8 = 800 kg m sˆ’1

Force exerted on the object, F = mv – mu / t

= m (v-u) / t

= 800 – 500

= 300 / 6

= 50 N

Initial momentum of the object is 500 kg m sˆ’1.

Final momentum of the object is 800 kg m sˆ’1.

Force exerted on the object is 50 N.

mark this as brainliest!

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3 years ago
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