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Deffense [45]
3 years ago
15

On the Moon's surface, lunar astronauts placed a corner reflector, off which a laser beam is periodically reflected. The distanc

e to the Moon is calculated from the round-trip time. The Earth's atmosphere slows down light. Assume the distance to the Moon is precisely 3.84×10^8 m, and Earth's atmosphere (which varies in density with altitude) is equivalent to a layer 40.0 km thick with a constant index of refraction n=1.000293. What is the difference in travel time for light that travels only through space to the moon and back and light that travels through the atmosphere and space?
Physics
1 answer:
Arisa [49]3 years ago
4 0

Answer:

Explanation:

The difference in time will be due to travel through atmosphere where speed of light slows down. If t be the thickness of atmosphere and c be the speed of light in space and μ be the refractive index of atmosphere difference in travel time will be as follows .

difference = \frac{2t\mu }{c}-\frac{2t }{c}

=\frac{2t}{c }\left ( 1-\mu  \right )

Now t = 40 x 10³m ,μ = 1.000293 , c = 3 x 10⁸.

difference =\frac{2t\mu }{c}-\frac{2t }{c}

=\frac{2t}{c }\left ( \mu -1  \right )\\

=\frac{ 2\times 40\times 10^3}{3\times10^3 }\left ( 1.000293-1 \right )\\

=7.81\times 10^{-3} s

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Using Rayleigh's criterion, what is the smallest separation between two pointlike objects that a person could clearly resolve at
atroni [7]

 Answer:

y = 52.44 10⁻⁶  m

Explanation:

It is Rayleigh's principle that two points are resolved if the maximum of the diffraction pattern of one matches the minimum the diffraction pattern of the other

Based on this principle we must find the angle of the first minimum of the diffraction expression

         a sin θ= m λ

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       sin θ =  λ  / a

Now let's use trigonometry the object is a distance L = 0.205 m

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Since the angles are very small, let's approximate

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We substitute in the diffraction equation

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7 0
3 years ago
A rescue plane travels horizontally at 30.0m/s 200.0m above the ground. it drops a supply package.
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2 years ago
Read 2 more answers
A photon with an energy E = 2.12 GeV creates a proton-antiproton pair in which the proton has a kinetic energy of 96.0 MeV. What
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The kinetic energy of the anti proton is 147.4 MeV.

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Using formula of energy

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We know that,

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So, E_{photon}=2mc^2+K.E_{p}+K.E_{np}

K.E_{np}=E_{photon}-(2mc^2+K.E_{p})

Put the value into the formula

K.E_{np}=2.12\times10^{9}-2\times938.3\times10^{6}-96\times10^{6}

K.E_{np}=147.4\ MeV

Hence, The kinetic energy of the anti proton is 147.4 MeV.

6 0
3 years ago
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