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gavmur [86]
2 years ago
13

For this problem, you will only be concerned with the geometric aspects of thin-film interference, so ignore phase shifts caused

by reflection from a medium with higher index of refraction. (Because of the structure of a butterfly's wings, such phase shifts do not contribute much to what you actually see when you look at the butterfly.) Part A Assume that light is incident normal to the surface of a film of thickness d. How much farther does the light reflected from the back surface travel than the light reflected from the front surface
Physics
1 answer:
Nana76 [90]2 years ago
4 0

Answer:

Λ = 2 t

Explanation:

This is a problem is interference by reflection, where it is requested not to take into account the changes of phase and change in the wavelength within the film

                   λₙ = λ₀ / n

When the wave enters the film, it must reach the end of the film, be reflected and reach the initial surface, therefore the total length is

                d = 2 t

where t is the thickness of the film, because the film is perpendicular to the surface

therefore the difference in optical path between the reflected ray on the surface and the end film is

                  Λ = 2 t

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lesantik [10]

Answer:

35870474.30504 m

Explanation:

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Radius of Earth = 6.38\times 10^6\ m

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\dfrac{GMm}{R^2}=m\dfrac{v^2}{R}\\\Rightarrow v=\sqrt{\dfrac{GM}{R}}

T=\dfrac{2\pi r}{v}\\\Rightarrow T=\dfrac{2\pi r}{\sqrt{\dfrac{GM}{r}}}

From Kepler's law we have relation

T^2=\dfrac{4\pi^2r^3}{GM}\\\Rightarrow r^3=\dfrac{T^2GM}{4\pi^2}\\\Rightarrow r=\left(\dfrac{(24\times 3600)^2\times 6.67\times 10^{-11}\times 5.98\times 10^{24}}{4\pi^2}\right)^{\dfrac{1}{3}}\\\Rightarrow r=42250474.30504\ m

Distance from the center of the Earth would be

42250474.30504-6.38\times 10^6=\mathbf{35870474.30504\ m}

8 0
3 years ago
Which of the following describes a role of gravity in the formation of our solar system?
Black_prince [1.1K]
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5 0
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pentagon [3]

Answer:

The well is 7.1 meters deep.

Explanation:

The formula to use here is the distance in a uniformly accelerated motion:

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Answer:

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