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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Read 2 more answers

A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm

Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

$F = - k * \Delta x (1)$

where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
Solving for k:

$k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)$

Assuming no friction present, total mechanical energy mus keep constant.
When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

$U = \frac{1}{2}* k* (\Delta x)^{2} (3)$

Replacing k and Δx by their values, we get:

$U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)$

When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
So, we can write the following expression:

$\frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2} = \frac{1}{2}*k*\Delta x^{2} (5)$