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rewona [7]
3 years ago
12

Which statements describe diffraction? Check all that apply.

Physics
2 answers:
Oksi-84 [34.3K]3 years ago
8 0

Answer: Diffraction occurs when waves interact with a gap.

Diffraction occurs in any type of wave—mechanical or electromagnetic.

Explanation:

The phenomenon of light to bend around a small obstruction in its path is known as diffraction. Diffraction occurs for any type of waves- transverse or longitudinal i.e. for mechanical or electromagnetic waves.

When the medium changes- refraction occurs. Waves reflect back from a barrier. Diffraction occurs when the wavelength of the wave is comparable to the size of the gap. The wave bends around the corners of the gap.

uranmaximum [27]3 years ago
6 0
<span>-- Diffraction occurs when waves interact with a barrier.
-- Diffraction occurs when waves interact with a gap.
-- Diffraction occurs in either type of wave—mechanical or electromagnetic.</span>
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Which best describes the transition from gas to liquid?

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Match the indices of refraction with the corresponding effects on the waves.
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Answer:

B) waves speed up

C) waves bend away from the normal

Explanation:

The index of refraction of a material is the ratio between the  speed of light in a vacuum and the speed of light in that medium:

n=\frac{c}{v}

where

c is the speed of light in a vacuum

v is the speed of light in the medium

We can re-arrange this equation as:

v=\frac{c}{n}

So from this we already see that if the index of refraction is lower, the speed of light in the medium will be higher, so one correct option is

B) waves speed up

Moreover, when light enters a medium bends according to Snell's Law:

n_1 sin \theta_1 = n_2 sin \theta_2

where

n_1 ,n_2 are the index of refraction of the 1st and 2nd medium

\theta_1,\theta_2are the angles made by the incident ray and refracted ray with the normal to the interface

We can rewrite the equation as

sin \theta_2 = \frac{n_1}{n_2}sin \theta_1

So we see that if the index of refraction of the second medium is lower (n_2), then the ratio \frac{n_1}{n_2} is larger than 1, so the angle of refraction is larger than the angle of incidence:

\theta_2>\theta_1

This means that the wave will bend away from the normal. So the other correct option is

C) waves bend away from the normal

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3 years ago
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The mean free path of a helium atom in helium gas at standard temperature and pressure is 0.2 um.What is the radius of the heliu
ivolga24 [154]

Answer: 0.10233nm

Explanation:

The mean free path \lambda   of an atom is given by the following formula:

\lambda=\frac{RT}{\sqrt{2} \pi d^{2}N_{A}P}    (1)

Where:

\lambda=0.2\mu m=0.2(10)^{-6}m

R=8.3145J/mol.K is the Universal gas constant

T=0\°C=273.115K is the absolute standard temperature

d is the diameter of the helium atom

N_{A}=6.0221(10)^{23}/mol is the Avogadro's number

P=1atm=101.3kPa=101.3(10)^{3}Pa=101.3(10)^{3}J/m^{3} absolute standard pressure

Knowing this, let's find d from (1), in order to find the radius r of the helium atom:

d=\sqrt{\frac{RT}{\sqrt{2}\pi\lambda N_{A}P}}    (2)

d=\sqrt{\frac{(8.3145J/mol.K)(273.115K)}{\sqrt{2}\pi(0.2(10)^{-6}m)(6.0221(10)^{23}/mol)(101.3(10)^{3}J/m^{3})}}    (3)

d=2.0467(10)^{-10}m    (4)

If the radius is half the diameter:

r=\frac{d}{2}  (5)

Then:

r=\frac{2.0467(10)^{-10}m}{2}  (6)

r=1.0233(10)^{-10}m  (7)

However, we were asked to find this radius in nanometers. Knowing 1nm=(10)^{-9}m:

r=1.0233(10)^{-10}m.\frac{1nm}{(10)^{-9}m}=0.10233nm  (8)

Finally:

r=0.10233nm This is the radius of the helium atom in nanometers.

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