All categories

3 years ago

What is cutoff wavelength?

Physics

1 answer:

LenKa [72]3 years ago

3 0

Answer:

The cutoff wavelength is the minimum wavelength in which a particular fiber still acts as a single mode fiber. Above the cutoff wavelength, the fiber will only allow the LP01 mode to propagate through the fiber (fiber is a single mode fiber at this wavelength).

Explanation:

You might be interested in

Which of the following statements describes the trend in the data?

AURORKA [14]

Answer:

Explanation:

The answer is D- The number of flies decreased, then leveled off.

6 0

2 years ago

Una furgoneta circula por una carretera a 55km/h. Diez km atrás , un coche circula en el mismo sentido a 85km/h ¿ En cuanto tiem

statuscvo [17]

Answer:

t = 0.33h = 1200s

x = 18.33 km

Explanation:

If the origin of coordinates is at the second car, you can write the following equations for both cars:

car 1:

$x=x_o+v_1t$ (1)

xo = 10 km

v1 = 55km/h

car 2:

$x'=v_2t$ (2)

v2 = 85km/h

For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:

$x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}$

$t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s$

The position in which both cars coincides is:

$x=(55km/h)(0.33h)=18.33km$

6 0

3 years ago

Which line represents a stationary object?

sineoko [7]

Answer:

I think its B but I may be wrong

3 0

3 years ago

Read 2 more answers

Andrea's near point is 20.0 cm and her far point is 2.0 m. Her contact lenses are designed so that she can see objects that are

Shalnov [3]

Answer:

the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

Explanation:

Given the data in the question,

near point = 20 cm

far point = 2 m = 200 cm

Now, for an object that is infinitely far away, the image is at is its far point.

so using the following expression, we can determine the focal length

1/f = 1/i + 1/o

where f is the focal length, i is the image distance and o is the object distance.

here, far point i = 2 m = 200 cm and v is ∞

so we substitute

1/f = 1/(-200 cm) + 1/∞

f = -200 cm

Also, for object at its closest point, the image appear at near point,

1/f = 1/i + 1/o

we make o the subject of formula

o = ( i × f ) / ( i - f )

given that near point i = 20 cm

we substitute

o = ( -20 × -200 ) / ( -20 - (-200) )

o = 4000 / 180

o = 22.2 cm

Therefore, the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

4 0

3 years ago

1. boiling point of water

melomori [17]

Answer:

what is the question. . .

5 0

3 years ago