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Lesechka [4]
3 years ago
10

How is the direction of light changed when it travels from an optically denser medium to an optically rarer medium????? please a

nswer it fast......​
Physics
1 answer:
ANTONII [103]3 years ago
4 0

Answer:

The light bends away from the normal

Explanation:

We can solve the problem by using Snell's law:

n_1 sin \theta_1 = n_2 sin \theta_2

where:

n_1 is the index of refraction of the first medium

n_2 is the index of refraction of the second medium

\theta_1 is the angle of incidence (angle between the incoming ray and the normal to the interface)

\theta_2 is the angle of refraction (angle between the outcoming ray and the normal to the interface)

We can rearrange the equation as

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

In this problem, light travels from an optically denser medium to an optically rarer medium, so

n_1 > n_2

Therefore, the term \frac{n_1}{n_2} is greater than 1, so

sin \theta_2 > sin \theta_1\\\rightarrow \theta_2 > \theta_1

which means that the angle of refraction is greater than the angle of incidence, and so the light will bend away from the normal.

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Answer:

2.5 kg.m/s

Explanation:

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Momentum for the ball moving towards the left side=mv=2.5*4=10 kg.m/s

Total momentum=-7.5 kg.m/s+10 kg.m/s=2.5 kg.m/s

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Fernanda is working on a physics project. she is experimenting by rolling a marble down a ramp and then seeing how far it rolls
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The distance covered on the floor after leaving the ramp is the dependent variable.

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A sophomore with nothing better to do adds heat to a mass 0.300 kg of ice at 0.0 âc until it is all melted.
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A proton is moving in a circular orbit of radius 12 cm in a uniform 0.31-T magnetic field perpendicular to the velocity of the p
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Answer:

Explanation:

A proton of charge

q=+1.609×10^-19C

Orbit a radius of 12cm

r=0.12m

Magnetic Field of 0.31T

Angle between velocity and field is 90°

a. Because the magnetic force F supplies the centripetal force Fc.

The magnitude of the magnetic force F on a charge q moving at a speed v in a magnetic field of strength B is given by

F = qvB sin θ

And the centripetal force is given as

Fc=mv²/r

Where m is mass of proton

m=1.673×10^-27kg

Then, F=Fc

qvB sin θ=mv²/r

qBSin90=mv/r

rqB=mv

Then, v=rqB/m

v=0.12×1.609×10^-19×0.31/1.673×10^-23

v=3577692.78m/s

v=3.58×10^6m/s

b. Since,

F=qVBSin90

F=1.609×10^-19×3.58×10^6×0.31

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