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3 years ago

How light is channelled down an optical fibre

Physics

1 answer:

coldgirl [10]3 years ago

3 0

Explanation:

Suppose you want to shine a flashlight beam down a long, straight hallway. Just point the beam straight down the hallway -- light travels in straight lines, so it is no problem. What if the hallway has a bend in it? You could place a mirror at the bend to reflect the light beam around the corner. What if the hallway is very winding with multiple bends? You might line the walls with mirrors and angle the beam so that it bounces from side-to-side all along the hallway. This is exactly what happens in an optical fiber.

The light in a fiber-optic cable travels through the core (hallway) by constantly bouncing from the cladding (mirror-lined walls), a principle called total internal reflection. Because the cladding does not absorb any light from the core, the light wave can travel great distances.

However, some of the light signal degrades within the fiber, mostly due to impurities in the glass. The extent that the signal degrades depends on the purity of the glass and the wavelength of the transmitted light (for example, 850 nm = 60 to 75 percent/km; 1,300 nm = 50 to 60 percent/km; 1,550 nm is greater than 50 percent/km). Some premium optical fibers show much less signal degradation -- less than 10 percent/km at 1,550 nm.

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Which statements describe solstices? Select 3 correct choices. They occur when the Sun reaches its highest or lowest point in th

Contact [7]

The statements that best describe solstices include:

A. They occur when the Sun reaches its highest or lowest point in the sky.

B. They affect the amount of sunlight in the Northern Hemisphere in June.

D. They influence the cycling of different seasons based on the Sun's height in the sky.

A solstice can be defined as a phenomenon in which the Sun reaches its maximum or minimum declination relative to the celestial equator on the celestial sphere, thereby, producing either the least amount of daylight time within a day or the most amount of daylight time. Also, it occurs twice in a year between June 20-22 and December 21-23.

Traditionally, solstice marks the beginning (start) of seasons such as summer and winter depending on the hemisphere.

Hence, the statements that best describe solstices include:

Solstices occur when the Sun reaches its highest (maximum) or lowest (minimum) point in the sky.

Solstices affect the amount of daylight time (sunlight) in the Northern Hemisphere in June.

Solstices influence the cycling of different seasons depending on the Sun's height in the sky.

Read more on solstices here: brainly.com/question/5133631

7 0

2 years ago

A box slides down a frictionless ramp.if it starts at rest, what is it’s speed at the bottom?

zhenek [66]

8.854 m/s is the speed of the box after it reaches bottom of the ramp.

<u>Explanation</u>:

From the figure we came to know that height of the block is 4 m.

We know that,

Total "initial energy of an object" = Total "final energy of an object "

Total "initial energy of an object" is = "sum of potential energy" and "kinetic energy" of an object at its initial position.

$\text { "g" acceleration due to gravity is } 9.8 \mathrm{m} / \mathrm{s}^{2}$

$\text { Total initial energy }=\mathrm{m} \times \mathrm{g} \times \mathrm{h}_{\mathrm{i}}+\frac{1}{2} \mathrm{m} v_{i}^{2}$

Initial velocity is “0” as the object does not have starting speed

$\text { Height of the block where the object is placed initially }\left(h_{i}\right) \text { is } 4 \mathrm{m} \text { . }$

$\text { Total initial energy }=\mathrm{m} \times 9.8 \times 4+\frac{1}{2} \mathrm{m} 0^{2}$

Total initial energy = 39.2 × m

$\text { Total final energy }=\mathrm{m} \times \mathrm{g} \times \mathrm{h}_{\mathrm{f}}+\frac{1}{2} m v_{f}^{2}$

$\text { We need to find final velocity } v_f$

$\text { Height of the block where the object is travelled to bottom (h_) is } 0 \mathrm{m} \text { . }$

$\text { Total final energy }=\mathrm{m} \times 9.8 \times 0+\frac{1}{2} m v_{f}^{2}$

Now, Total initial energy of an object = Total final energy of an object

$39.2 \times \mathrm{m}=0.5 \mathrm{m} v_{f}^{2}$

$\frac{39.2}{0.5}=v_{f}^{2}$

$v_{f}^{2}=78.4$

$v_{f}=\sqrt{78.4}$

$v_{f}=8.854 \mathrm{m} / \mathrm{s}$

Final speed is 8.854 m/s.

3 0

3 years ago

What Is the acceleration of a toy car that starts from rest and has a speed of 0.12 m/s after 0.1?

VikaD [51]

Given: Initial velocity of toy car (u ) = 0

Final velocity of toy car (v) = 0.12 m/s

Required time (t) = 0.1 s

To find: The acceleration of the toy car.

Let the acceleration of the toy car be (a)

Formula Used: 1st kinematic equation of motion

v = u + at ---------------------------(i)

Here, all alphabets are in their usual meanings

Now, from equation (i), we shall calculate the value of 'a'.

so, a = (v - u) /t

or, a = (0.12 m/s - 0) / 0.1s

or, a = 1.2 m/s²

Hence, the required acceleration of the toy car will be 1.2 m/s².

6 0

3 years ago

An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 49.8 N,

tatiyna

Answer:

W = 7.06 J

Explanation:

From the given information the spring constant 'k' can be calculated using the Hooke's Law.

$F = kx\\49.8 = k(0.181)\\k = 275.13~N/m$

Now, using this spring constant the additional work required by F to stretch the spring can be found.

The work energy theorem tells us that the work done on the spring is equal to the change in the energy. Therefore,

$W = U_2 - U_1\\W = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2 = \frac{1}{2}(275.13)[0.29^2 - 0.18^2] = 7.06~J$

6 0

3 years ago

The department of insurance and safety is led by what official

lawyer [7]

Answer: Georgia Department of Insurance

Explanation: I hope this help :]

6 0

3 years ago