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4 years ago

Two or more waves combining to produce a wave with a smaller displacement is called _____.

2 answers:

6 0

<u><em>Answer:</em></u>
Destructive interference

<u><em>Explanation:</em></u>
<u>Interference</u> is a term used to refer to the combination of two waves or their overlap
<u>There are two types of interference:</u>
<u>1- Constructive interference:</u> This happens when the waves overlap/combine to produce a resultant wave with a larger displacement
<u>2- Destructive interference:</u> This happens when the waves overlap/combine to produce a resultant wave with a smaller displacement

Hope this helps :)

Bumek [7]4 years ago

6 0

Two or more waves combining to produce a wave with a smaller displacement is called Destructive interference.

Explanation:

The interference of 2 waves of equal frequency and opposite section, leading to their cancellation wherever the negative displacement of 1 perpetually coincides with the positive displacement of the opposite.

In harmful interference, the crest of 1 wave meets the trough of another, and therefore the result's a lower total amplitude.

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I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi

Masteriza [31]

Answer:

After it has traveled through 1 cm : $I(1 \ cm) = 0.5488 I_0$

After it has traveled through 2 cm : $I(2 \ cm) = 0.3012 I_0$

After it has traveled through 1 cm : $od( 1\ cm) = 0.2606$

After it has traveled through 2 cm : $od( 2\ cm) = 0.5211$

Explanation:

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

The optical density od is given by:

$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$ .

So, after travelling 1 cm:

$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$

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