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frozen [14]
3 years ago
14

When light propagates through two adjacent materials that have different optical properties, some interesting phenomena occur at

the interface separating the two materials. For example, consider a ray of light that travels from air into the water of a lake. As the ray strikes the air-water interface (the surface of the lake), it is partly reflected back into the air and partly refracted or transmitted into the water. This explains why on the surface of a lake sometimes you see the reflection of the surrounding landscape and other times the underwater vegetation.
These effects on light propagation occur because light travels at different speeds depending on the medium. The index of refraction of a material, denoted by n , gives an indication of the speed of light in the material. It is defined as the ratio of the speed of light c in vacuum to the speed v in the material, orn=cv

What is the minimum value that the index of refraction can have?
Physics
1 answer:
Veseljchak [2.6K]3 years ago
4 0

Answer:

First, the different indices of refraction must be taken into account (in different media): for example, the refractive index of light in a vacuum is 1 (since vacuum = c).  The value of the refractive index of the medium is a measure of its "optical density":  Light spreads at maximum speed in a vacuum but slower in others  transparent media; therefore in all of them n> 1. Examples of typical values ​​of  are those of air (1,0003), water (1.33), glass (1.46 - 1.66) or diamond (2.42).

The refractive index has a maximum value and a minimum value, which we can calculate the minimum value by means of the following explanation:

The limit or minimum angle, α lim, is defined as the angle of refraction from which  the refracted ray disappears and all the light is reflected. As in the maximum value of  angle of refraction, from which everything is reflected, is βmax = 90º, we can  know the limit angle (the minimum angle that we would have to have to know the minimum index of refraction) by Snell's law:

 βmax = 90º ⇒ n 1x sin α (lim) = n 2 ⇒ sin α lim = n 2 / n 1

Explanation:

When a light ray strikes the separation surface between two media  different, the incident beam is divided into three: the most intense penetrates the second  half forming the refracted ray, another is reflected on the surface and the third is  breaks down into numerous weak beams emerging from the point of incidence in  all directions, forming a set of stray light beams.

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Answer:

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he average or effective voltage of a wave is the value of the wave in a period

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in this case the given volage is a square wave that can be described by the function

           V (t) = \left \{ {{V=V_o \ \ \  t<  \tau /2} \atop {V=0 \ \  \ \  t> \tau /2 }   } \right.

to substitute in the equation let us separate the into two pairs

             V_average = \int\limits^{1/2}_0 {V_o} \, dt + \int\limits^1_{1/2} {0} \, dt

             V_average = V_o \ \int\limits^{1/2}_0 {} \, dt

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we evaluate  V₀ = 4 V

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Answer:

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Substituting the given parameters into the formula

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