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weeeeeb [17]
3 years ago
11

A ray of light passes from air into a block of clear plastic. How does the angle of incidence in the air compare to the angle of

refraction in the plastic? A ray of light passes from air into a block of clear plastic. How does the angle of incidence in the air compare to the angle of refraction in the plastic? The angle of incidence is equal to the angle of refraction. The angle of incidence is greater than the angle of refraction. The angle of incidence is less than the angle of refraction. The two angles cannot be compared without additional information.
Physics
1 answer:
andre [41]3 years ago
6 0

Answer:

The angle of incidence is greater than the angle of refraction

Explanation:

Refraction occurs when a light wave passes through the boundary between two mediums.

When a ray of light is refracted, it changes speed and direction, according to Snell's Law:

n_1 sin \theta_1 = n_2 sin \theta_2

where :

n_1 is the index of refraction of the 1st medium

n_2 is the index of refraction of the 2nd medium

\theta_1 is the angle of incidence (the angle between the incident ray and the normal to the boundary)

\theta_2 is the angle of refraction (the angle between the refracted ray and the normal to the boundary)

In this problem, we have a ray of light passing from air into clear plastic. We have:

n_1=1.00 (index of refraction of air)

n_2=1.50 approx. (index of refraction in clear plastic)

Snell's Law can be rewritten as

sin \theta_2 =\frac{n_1}{n_2}sin \theta_1

And since n_2>n_1, we have

\frac{n_1}{n_2}

And so

\theta_2

Which means that

The angle of incidence is greater than the angle of refraction

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AnnyKZ [126]

Answer:

Fy=107.2 N

Explanation:

Conceptual analysis

For a right triangle :

sinβ = y/h formula (1)

cosβ = x/h formula (2)

x: side adjacent to the β angle

y:  opposite side of the β angle

h: hypotenuse

Known data

h = T = 153.8 N : rope tension

β= 44.2°with the horizontal (x)

Problem development

We apply the formula (1) to calculate Ty : vertical component of the rope force.

sin44.2° = Ty/153.8 N

Ty = (153.8 N ) *(sen44.2°)= 107.2 N  directed down

for equilibrium system

Fy= Ty=107.2 N

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3 years ago
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vredina [299]

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3 years ago
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A particle cannot generally be localized to distances much smaller than its de Broglie wavelength. This fact can be taken to mea
Vlada [557]

The De broglie wavelength of a thermal neutron at room temperature 300K = 1.5 × A°

<h3>How is the De broglie wavelength of a thermal neutron at room temperature calculated?</h3>

Temperature, T = 300K

Momentum, p = mv

Therefore v = p/m

Energy, E= 1/2 m( p/m) ²

Boltzman Energy= 3/2 KT

3/2KT = 1/2 m(p/m)²

Therefore p =  \sqrt{3RTm}

According to De broglie hypothesis, P = h ÷ λ

Therefore,    λ  = h ÷ \sqrt{3RTm}

                     = 6.6× 10^{-34} ÷  \sqrt{3 \times 1.38 \times  10^{-23}  \times 300 \times 1.6 \times  10^{-27} }

                     = 0.15 × 10^{-9}

Therefore the De broglie wavelength of a thermal neutron at room temperature 300K = 1.5 × A°

To learn more about De broglie wavelength, refer: <u>https://brainly.in/question/6131028</u>

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1 year ago
Concept Development
Ostrovityanka [42]

Answer:

The answer is below

Explanation:

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A magnifier has a magnification of 8 x. How far from the lens should an object be placed so that it’s virtual image is at the ne
larisa86 [58]

The distance of the object from the lens is 3.12 cm.

<h3>What is magnification?</h3>

Magnification is the process of of enlarging an apparent size of an object.

<h3>Object distance</h3>

The object distance is calculated using the following lens formula;

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where;

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Thus, the distance of the object from the lens is 3.12 cm.

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