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Gekata [30.6K]
2 years ago
6

In the shadow of a tree with a dense, leafy canopy, one sees a number of light spots. Surprisingly, they all appear to be circul

ar. Why
Physics
1 answer:
Bad White [126]2 years ago
8 0

The characteristics of the diffraction phenomenon allow to find the result for the shape of the points of light that you pass the tree is:

  • The shape of the dots is circular because it is in the range of far-field diffraction.

Diffraction is the phenomenon where the undulatory part of the light becomes evident, it is the interference of the waves that make up each ray of light, for this phenomenon to occur it must be fulfilled that the wavelength is of the order of the space where pass the light.

In the leafy tree it has many leaves, but there are spaces between them, some of these spaces are small and it fulfills the diffraction condition, therefore we see bright spots and not a continuous shadow.

Diffraction can be classified depending on the distance to the observer:

  • Near field or fresnel. In this case the distance from the observer is small and we can see the shape of the object that creates the diffraction.
  • Far field or Fraunhoger. In this case the distance between the obstacle (leaves) and the person is great, here the information on the shape of things is lost and we have two observable forms. Lines for the case of slits and circles for the case of objects with a closed shape.

In this case, the distance from the leaves to the observer is large, therefore we are in the case of far-field diffraction and since the edge of the leaves that forms the diffraction is closed, the observable shape is a circle.

In conclusion using the characteristics of the diffraction phenomenon we can find the result for the shape of the points of light that pass the tree is:

  • The shape of the dots is circular because it is in the range of far-field diffraction.

Learn more about diffraction here:  brainly.com/question/20140459

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Imagine you want to create a model of Solar system (and beyond) with the Sun
kondaur [170]

Using the scale model of the Sun given;

  • The diameter of the Milky Way = 4.28 × 10¹¹ m
  • The diameter of the Earth = 5.44 × 10⁻³ m

<h3>What is the diameter of the Milky Way?</h3>

The diameter of the Milky Way is about 1 × 10¹⁸ km.

The diameter of the Sun is about 1.4 × 10⁶ km

The diameter of the Earth is about 1.27 × 10⁴ km.

Using the scale model of the Sun given, the diameter of the Milky Way = (1 × 10¹⁸ km/1.4 × 10⁶ km) × 0.6 m

The diameter of the Milky Way = 4.28 × 10¹¹ m

Using the scale model of the Sun given, the diameter of the Milky Way = (1.27 × 10⁴ km/1.4 × 10⁶ km) × 0.6 m

The diameter of the Earth = 5.44 × 10⁻³ m

In conclusion, the diameter of the Milky Way is far bigger than the Sun while the diameter of the Sun is about 5400 times bigger than the Earth.

Learn more about the Milky Way, Sun, and Earth at: brainly.com/question/1995133

#SPJ1

5 0
2 years ago
The _______ is the layer of the atmosphere in which weather occurs
seraphim [82]
The "Troposphere" <span>is the layer of the atmosphere in which weather occurs!

It is the lowermost layer of atmosphere, where we live.

Hope this helps!</span>
8 0
3 years ago
Sort the properties of metalloids into the correct categories
enot [183]

Boron

Sillicon

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Arsenic

Antimony

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3 0
4 years ago
A spiral staircase winds up to the top of a tower in an old castle. To measure the height of the tower, a rope is attached to th
boyakko [2]

Answer:

8.76762 m

Explanation:

T = Time period = 5.94 seconds

g = Acceleration due to gravity = 9.81 m/s²

L = Length of pendulum = Height of tower

Time period is given by

T=2\pi \sqrt{\dfrac{L}{g}}\\\Rightarrow T^2=4\pi^2\dfrac{L}{g}\\\Rightarrow L=\dfrac{T^2g}{4\pi^2}\\\Rightarrow L=\dfrac{5.94^2\times 9.81}{4\pi^2}\\\Rightarrow L=8.76762\ m

The height of the tower is 8.76762 m

8 0
3 years ago
Water flows past a flat plate that is oriented parallel to the flow with an upstream velocity of 0.4 m/s. (a) Determine the appr
Trava [24]

Answer:

1.12 m

0.08291 m

Explanation:

u = Upstream velocity = 0.4 m/s

Re = Reynold's number = 5\times 10^5 (turbulent)

\nu = Viscosity of water = 1.12\times 10^{-6}\ Pas

Here the flow is turbulent so we have the relation

Re_{xcr}=\frac{ux_{cr}}{\nu}\\\Rightarrow x_{cr}=\frac{Re_{xcr}\nu}{u}\\\Rightarrow x_{cr}=\frac{5\times 10^5\times 1.12\times 10^{-6}}{0.4}\\\Rightarrow x_{cr}=1.4\ m

The approximate location downstream from the leading edge where the boundary layer becomes turbulent is 1.4 m

Boundary layer thickness relation is given by

\delta={\frac{\nu x}{u}}^{\frac{1}{5}}\\\Rightarrow \delta={\frac{1.12\times 10^{-6}\times 1.4}{0.4}}^{\frac{1}{5}}\\\Rightarrow \delta=0.08291\ m

The boundary layer thickness is 0.08291 m

4 0
4 years ago
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