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kramer
3 years ago
8

Light incident on a lake surface is partly reflected and partly refracted.What is the differences between the reflected ray and

incident ray?
Physics
2 answers:
iren2701 [21]3 years ago
5 0

Answer: As per the question, a ray of light is incident on a surface and it is partly reflected and refracted. The incident light is an unpolarised light. The reflected light is partially polarised.

If the angle of incidence becomes equal to the Brester angle (polarising angle), then the reflected light is completely plane polarised.

nika2105 [10]3 years ago
5 0
The incident ray is a ray that hits the surface of the water. A reflected ray will always correspond to the incident ray and it is the light that is reflected by the surface of the water. Meaning, all rays that hit are incident rays, however, some are reflected rays.
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Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
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a.\rm -1.49\ m/s^2.

b. \rm 50.49\ m.

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<u>Given:</u>

  • Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

<h2>(a):</h2>

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).

At time t = 3 seconds,

\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.

For the time interval of 2 seconds,

\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.

It is the displacement of the particle in 2 seconds.

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