Two in-phase sources of waves are separated by a distance of 4.00 m. These sources produce identical waves that have a wavelength of 5.00 m. On the line between them, there are two places at which the same type of interference occurs. (a) Is it constructive or destructive interference, and (b) where are the places located

It's a Destructive interference because of the fact that The wavelength is 5m while the sources are separated by 4m. Constructive interference could only have been possible if and when the path difference is 5.

b)

Assuming that m = 0, then

Δl = (0 + ½)λ

Δl = ½λ, where Δl = l2-l1

l2 - l1 = ½λ

l2 - l1 = 5/2

l2 - l1 = 2.5

On the other hand, the question says that l1 + l2 = 4. So, on solving simultaneously, we have

l2 + l1 = 4

l2 - l1 = 2.5

2l1 = 1.5

l1 = 1.5/2

l1 = 0.75 m

l2 = 4 - l1

l2 = 4 - 0.75

l2 = 3.25 m

shubhamchouhanvrVT shubhamchouhanvrVT · Answer 2 · 2022-04-12T13:00:13+02:00

a) Two in-phase sources of waves are separated by a distance will have destructive interference

b) The location will be 0.75 m and 3.25 m

What is destructive interference?

Destructive interference occurs when the maxima of two waves are 180 degrees out of phase: a positive displacement of one wave is cancelled exactly by a negative displacement of the other wave. The amplitude of the resulting wave is zero.

Taking a look at constructing interference, the path difference is given as

Δl = mλ

On the other hand, path difference for its destructive interference is given as

Δl = (m + ½) λ

a) It's a Destructive interference because of the fact that The wavelength is 5m while the sources are separated by 4m. Constructive interference could only have been possible if and when the path difference is 5.

b)Assuming that m = 0, then

Δl = (0 + ½)λ

Δl = ½λ, where Δl = l2-l1

l2 - l1 = ½λ

l2 - l1 = 5/2

l2 - l1 = 2.5

On the other hand, the question says that [tex]I_1+I_2=4[/tex] . So, on solving simultaneously, we have

[tex]l_2 + l_1 = 4[/tex]

[tex]l_2 - l_1 = 2.5[/tex]

[tex]2 \ l_1 = 1.5[/tex]

[tex]l_1 = \dfrac{1.5}{2}[/tex]

[tex]l_1 = 0.75 m[/tex]

[tex]l_2 = 4 - l_1[/tex]

[tex]l_2 = 4 - 0.75[/tex]

[tex]l_2 = 3.25 m[/tex]

Hence the location will be 0.75 m and 3.25 m

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