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Deffense [45]
4 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]4 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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A convex thin lens with refractive index of 1.50 has a focal length of 30cm in air. When immersed in a certain transparent liqui
GalinKa [24]

Answer:

n_l = 1.97

Explanation:

given data:

refractive index of lens 1.50

focal length in air is 30 cm

focal length in water is -188 cm

Focal length of lens is given as

\frac{1}{f} =\frac{n_2 -n_1}{n_1} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

\frac{1}{f} =\frac{n_{g} -n_{air}}{n_{air}} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

\frac{1}{f} =\frac{n_{g} -1}{1} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

focal length of lens in liquid is

\frac{1}{f} =\frac{n_{g} -n_{l}}{n_{l}} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

                =\frac{n_{g} -n_{l}}{n_{l}}  [\frac{1}{(n_{g} - 1) f}

rearrange fron_l

n_l = \frac{n_g f_l}{f_l+f(n_g-1)}

n_l = \frac{1.50*(-188)}{-188 + 30(1.50 -1)}

n_l = 1.97

7 0
3 years ago
2. True or False Every sample of a pure substance has exactly the same composition
Marrrta [24]

Answer:

the correct answer is False

Explanation:

i hope its right

5 0
3 years ago
If E=1/2Av^2+Bmx find the dimensions of A and B(Where E is energy,v,m and x are velocity,mass and distance respectively)
sp2606 [1]

Answer:

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Explanation:

E = ½ Av² + Bmx

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A Joule written in base units is:

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Each term must have the same units.

[kg m²/s²] = A [m/s]²

[kg m²/s²] = A [m²/s²]

A = [kg]

[kg m²/s²] = B [kg] [m]

B = [m/s²]

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Answer:

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