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Slav-nsk [51]
3 years ago
15

Two light waves are initially in phase and have the same wavelength, 470 nm. They enter two different media of identical lengths

of 2.50 µm. If n1 = 1.2 and n2 = 1.5, what is the effective phase difference of the waves when they exit the media? Group of answer choices 3.7 rad 10 rad 1.6 rad 0.6 rad 0 rad
Physics
1 answer:
-Dominant- [34]3 years ago
5 0

Answer:

= 3.7 radians = effective phase difference

Explanation:

Wavelength given =\lambda= 470 nm

n1 = 1.2 and n2 = 1.5

The light rays are traveling at different speeds inside the 2 different layers.

So effective optical path is given by  n d

where n is the refractive index of the medium and

d is the length traveled by the light ray in the medium.

The path difference introduced because of 2 different medium is

length = distance traveled by the wave = d = 2.5 x 10^-6 m

Phase difference =2 \pi / \lambda * path difference

path difference = (n₂ - n₁ ) d

= (1.5 - 1.2) (2.5 x 10⁻⁶)

= 7.5 x 10⁻⁷ m

phase difference = 2 pi / (470 x 10⁻⁹) * (7.5 x 10⁻⁷) = 10 rad

2 pi radians is 360 degrees

10 radians = 573 degrees

Phase difference = 573 - 360 = 212 degrees

= 3.7 radians = effective phase difference

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This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

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Explanation:

Given that;

∑'MA = 0

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we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

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NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

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we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

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FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

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