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snow_lady [41]
3 years ago
14

How does the bending of light depend on the two media?

Physics
1 answer:
erica [24]3 years ago
5 0

Answer:

Explanation:

The two media must have differing index of refraction.

Index of refraction is an indication of how fast light can move through the media.

If a light wave approaches an interface at an angle and the media the light is moving into has a higher index of refraction meaning slower light speed in the media, then the part of the wave front hitting first gets slowed down sooner. As each section of the wave front crosses the interface, the whole wave front has changed direction toward the side with the first drag. Much like a column of marching soldiers executing a slight turn each at the precise moment to keep both columns and lined up in "military precision"

If the media the light is moving into has a smaller index of refraction, meaning higher light speed, then the side of the light beam hitting the interface first speeds up before the opposite side of the beam. This makes the angle leaving the interface much closer to the interface surface than the angle inside the higher index media. There becomes a point called the critical angle where the light cannot exit the higher index of refraction and the condition of total internal reflection exists. Think fiber optic cable which can transmit signals hundreds of miles without significant loss.

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Answer:

speed of light simulating traveling at the speed of light. Speed of light, speed at which light waves propagate through different materials. In particular, the value for the speed of light in a vacuum is now defined as exactly 299,792,458 metres per second

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3 years ago
Atoms of the same element that differ only in the number of neutrons are known as.
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They are known as isotopes

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2 years ago
A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re
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The box has 3 forces acting on it:

• its own weight (magnitude <em>w</em>, pointing downward)

• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)

• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>

• net perpendicular force:

∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

<em>n</em> = <em>w</em> cos(35°)

<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)

<em>n</em> ≈ 120 N

Solve for the mag. of friction:

<em>f</em> = <em>µ</em> <em>n</em>

<em>f</em> = 0.25 (120 N)

<em>f</em> ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>

<em>a</em> ≈ (54.3157 N) / (15 kg)

<em>a</em> ≈ 3.6 m/s²

Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:

<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>

<em>v</em>² = 2 (3.6 m/s²) (3 m)

<em>v</em> = √(21.7263 m²/s²)

<em>v</em> ≈ 4.7 m/s

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<u>Answer:</u>

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Norma-Jean [14]
Im going to go with a. True
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