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Tanya [424]
3 years ago
11

What property of light waves does the michelson-morley interferometer directly demonstrate?

Physics
1 answer:
yan [13]3 years ago
8 0
The wave nature of light, due to the experiment having bright and dark bands corresponding to places where you have constructive and destructive interference.
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Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the pow
allsm [11]

This question is incomplete, the complete question;

you make an interferometer using 50-50 beam splitter and two mirrors, one being a perfect mirror and one which does not reflect all light. The wavelength of the 9 mW incident laser is 400 nm.

Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the power measured at the detector when only the vertical arm is blocked is 2.25 mW, while the power measured at the detector when only the horizontal arm is blocked is only 2.025 mW. Assume initially the intensity is at its maximum. How much would we need to translate the perfect mirror to the right to get a minimum intensity at detector, and what is that minimum intensity

Options;

a) 200 nm; 0.9 mW

b) 100 nm, 0.0059 mW

c) 200 nm; 0 mW

d) 100 nm; 0.9 mW

e) 200 nm; 0.0059 mW

Answer:

the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector  and the minimum intensity are;

100 nm; 0.0059 mW

Option b) 100 nm, 0.0059 mW is the correct answer

Explanation:

Given that the instrument here is an interferometer.

Maximum intensity is obtained when the two waves are exactly in phase.

that is the peaks (crusts and troughs) and nodes (zero value points) of the two waves will be at the exact same point when the wave falls on the detector.

The phase factor of this point is taken as ∅ = 0

Now, to get a minimum point, the phase difference between the two waves should be should be ∅ = π

This corresponds to a path difference between the two waves as half of the wavelength. λ/2

The light gets reflected from the mirror.

Hence, when we move the mirror by a length l, the extra/less path the light has to travel is 2l (light is going and coming back)

hence, to get a path difference of λ/2 the mirror should move half of this distance only

so, the mirror should move;

l = λ/4

here, wavelength is 400nm

the length moved by the mirror = 400/4 = 100 nm

The intensity is given by the equation;

l = l1 + l2 + 2√l1l2cos(∅)

where

l1 = 2.25 mW

l2 = 2.025 mW

∅ = π

so we substitute

l = 2.25 + 2.025 - 2√(2.25 × 2.025)

l = 4.275 - 4.26907

l = 0.0059

Therefore; the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector  and the minimum intensity are;

100 nm; 0.0059 mW

Option b) 100 nm, 0.0059 mW is the correct answer  

5 0
2 years ago
What's the difference between mechanical waves and vibrations
OlgaM077 [116]

A mechanical wave is a wave that is an oscillation of matter, and therefore transfers energy through a medium. While waves can move over long distances, the movement of the medium of transmission—the material—is limited. Therefore, the oscillating material does not move far from its initial equilibrium position.

7 0
3 years ago
What is the resistance if the current is 15 amps and the voltage 105 volts
Ludmilka [50]
I think it’s R=U/I = 120/5 = 24
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3 years ago
Summarize the weather conditions related to a warm front. Remember to include all data collected on warm fronts in this activity
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Answer:

to a warm front. Remember to include all data collected on warm fron … ... Remember to include all data collected on warm fronts in this activity to support your answer (examples: interaction of air masses, air pressure, cloud cover, temperature behind/ahead of front, wind direction, precipitation, etc. 1

Explanation:

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3 years ago
In preparation for building a space station, an astronaut removes a self-telescoping uniform rod from the cargo bay and releases
emmainna [20.7K]

To solve this problem it is necessary to apply the concepts related to the conservation of angular momentum. This can be expressed mathematically as a function of inertia and angular velocity, that is:

L = I\omega

Where,

I = Moment of Inertia

\omega= Angular Velocity

For the given object the moment of inertia is equivalent to

I = \frac{mr^2}{12}

Considering that the moment of inertia varies according to distance, and that there are two of these without altering the mass we will finally have to

L_i = L_f

I_i \omega_1 = I_f \omega_2

(\frac{mr_{initial}^2}{12})(\omega_1)=(\frac{mr_{final}^2}{12})(\omega_2)

(r_{initial}^2})(\omega_1)=(r_{final}^2)(\omega_2)

Our values are given as,

r_{initial} = 3m\\\omega_1 = 0.05rad/s \\r_{final}=1.5m

Replacing we have,

(3^2})(0.05)=(1.5^2)(\omega_2)

\omega_2 = 0.2rad/s

Therefore the angular speed after the catch slips is 0.2rad/s

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2 years ago
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