A group of protons and neutrons surrounded by electrons
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.
Galaxies are sprawling systems of dust, gas, dark matter, and anywhere from a million to a trillion stars that are held together by gravity. Nearly all large galaxies are thought to also contain supermassive black holes at their centers.
The atom is the most basic unit of matter
To solve this problem we will apply the concepts related to Ohm's law and Electric Power. By Ohm's law we know that resistance is equivalent to,
Here,
V = Voltage
I = Current
While the power is equivalent to the product between the current and the voltage, thus solving for the current we have,
Applying Ohm's law
Therefore the equivalent resistance of the light string is 
