After one meter, 3.4% of the light is gone ... either soaked up in the fiber
material or escaped from it. So only (100 - 3.4) = 96.6% of the light
remains, to go on to the next meter.
After the second meter, 96.6% of what entered it emerges from it, and
that's 96.6% of 96.6% of the original signal that entered the beginning
of the fiber.
==> After 2 meters, the intensity has dwindled to (0.966)² of its original level.
It's that exponent of ' 2 ' that corresponds to the number of meters that the light
has traveled through.
==> After 'x' meters of fiber, the remaininglight intensity is (0.966) ^x-power
of its original value.
If you shine 1,500 lumens into the front of the fiber, then after 'x' meters of
cable, you'll have
<em>(1,500) · (0.966)^x</em>
lumens of light remaining.
=========================================
The genius engineers in the fiber design industry would not handle it this way.
When they look up the 'attenuation' of the cable in the fiber manufacturer's
catalog, it would say "15dB per 100 meters".
What does that mean ? Break it down: 15dB in 100 meters is <u>0.15dB per meter</u>.
Now, watch this:
Up at the top, the problem told us that the loss in 1 meter is 3.4% . We applied
super high mathematics to that and calculated that 96.6% remains, or 0.966.
Look at this ==> 10 log(0.966) = <em><u>-0.15</u> </em> <== loss per meter, in dB .
Armed with this information, the engineer ... calculating the loss in 'x' meters of
fiber cable, doesn't have to mess with raising numbers to powers. All he has to
do is say ...
-- 0.15 dB loss per meter
-- 'x' meters of cable
-- 0.15x dB of loss.
If 'x' happens to be, say, 72 meters, then the loss is (72) (0.15) = 10.8 dB .
and 10 ^ (-10.8/10) = 10 ^ -1.08 = 0.083 = <em>8.3%</em> <== <u>That's</u> how much light
he'll have left after 72 meters, and all he had to do was a simple multiplication.
Sorry. Didn't mean to ramble on. But I do stuff like this every day.
Voltage = 1.5V
Resistance = 1.5 ohm
Current = V/R = 1.5/1.5 = 1A
A natural system is formed when components are grouped together so as to perform a more specific function or use. In addition, this system is bound to exist without any human interference. Biological and non-biological components work hand-in-hand as they do nature processes.
Explanation:
1. The Centauri star system consists of 3 stars and is about 4.37 light-years away from Earth. (A light-year is the distance that light travels in a year.)
2. We see the visible spectrum which consists of colors red, orange, yellow, green, blue, indigo, and violet. (The visible spectrum is the range of light that our eyes can detect.)
3. Because air has a lower index of refraction than water, the light beams making up the image of the fish I see bends as it leaves the water and enters the air. (Index of refraction determines how light bends when moving from one medium into another.)
4. When I look at my reflection in the mirror, the light reflects off the mirror at an equal but opposite angle as it hits the mirror.
5. When looking into a concave mirror the center is further away from you than the outside edges. (Concave mirrors cave in.)
6. Forklift drivers are careful to look in the convex mirrors placed at blind intersections in the warehouse.
7. is transparent and the light goes through without distortion.
8. The electromagnetic spectrum refers to any or all of the possible wavelengths of "light", both the visible light and invisible light.
9. The transmission of light through a medium means it goes through it.
10. The absorption of light by a medium means it is absorbed by it.
Explanation:
Effective nuclear charge is defined as he net positive charge experienced by an electron in an atom. It is termed "effective" because the shielding effect of electrons prevents higher orbital electrons from experiencing the full nuclear charge of the nucleus due to the repelling effect of inner-layer electrons.
The 1s is the closest shell to the nucleus of an therefore maximum nuclear charge is experienced. The formula for effective nuclear charge is:
Zeff = Z – S
where
Z = the number of protons in the nucleus, and
S = the shielding constant, the average number of electrons between the nucleus and the electron.
Hence, the energy required to remove an electron from the 1s orbital is the strongest.
