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fgiga [73]
3 years ago
9

What is the bouncing of light when it changes media

Physics
1 answer:
matrenka [14]3 years ago
5 0

When light passes from one medium to another, part of it continues on
into the new medium, while the rest of it bounces away from the boundary,
back into the first medium.

The part of the light that continues on into the new medium is <em>transmitted</em>
light.  Its forward progress at any point in its journey is <em>transmission</em>. 

Its direction usually changes as it crosses the boundary.  The bending is <em>
refraction</em>
.

The part of the light that bounces away from the boundary and heads back
into the first medium is <em>reflected</em> light.  The process of bouncing is <em>reflection</em>.


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A Huge water tank is 2m above the ground if the water level on it is 4.9m high and a small opening is there at the bottom then t
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Answer:

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

Explanation:

Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure (P_{atm}), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:

P_{1} + \rho\cdot \frac{v_{1}^{2}}{2} + \rho\cdot g \cdot z_{1} = P_{2} + \rho\cdot \frac{v_{2}^{2}}{2} + \rho\cdot g \cdot z_{2}

Where:

P_{1}, P_{2} - Water total pressures inside the tank and at ground level, measured in pascals.

\rho - Water density, measured in kilograms per cubic meter.

g - Gravitational acceleration, measured in meters per square second.

v_{1}, v_{2} - Water speeds inside the tank and at the ground level, measured in meters per second.

z_{1}, z_{2} - Heights of the tank and ground level, measured in meters.

Given that P_{1} = P_{2} = P_{atm}, \rho = 1000\,\frac{kg}{m^{3}}, g = 9.807\,\frac{m}{s^{2}}, v_{1} = 0\,\frac{m}{s}, z_{1} = 6.9\,m and z_{2} = 4.9\,m, the expression is reduced to this:

\left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m) = \frac{v_{2}^{2}}{2} + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.9\,m)

And final speed is now calculated after clearing it:

v_{2} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m-4.9\,m)}

v_{2} \approx 6.263\,\frac{m}{s}

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

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