# phyiscs help needed!!

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Could someone pls explain to me how to solve phase difference and path difference questions?

I have no idea

I have no idea

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#3

https://physicsthings.wordpress.com/...-difference-2/

https://www.thestudentroom.co.uk/sho....php?t=1664097

Looks helpful...

https://www.thestudentroom.co.uk/sho....php?t=1664097

Looks helpful...

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I don't understand how to solve this or questions like this. COuld you help me!

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#5

(Original post by

I don't understand how to solve this or questions like this. COuld you help me!

**ThirstyLearner**)I don't understand how to solve this or questions like this. COuld you help me!

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(Original post by

Trying...

**Spannerin'moi**)Trying...

With the formula of 2pi times path difference divided by wavelength gives 3pi

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#7

(Original post by

the answer is C, but idk how.

With the formula of 2pi times path difference divided by wavelength gives 3pi

**ThirstyLearner**)the answer is C, but idk how.

With the formula of 2pi times path difference divided by wavelength gives 3pi

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(Original post by

I see..was the path difference 0?

**Spannerin'moi**)I see..was the path difference 0?

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#9

(Original post by

Mark scheme says- π radians is the phase difference between the two waves at A – the path difference between the two waves is (8cm – 5cm) = 3cm = 1.5λ. This results in the waves arriving in antiphase (π radians out of phase) which also occurs when the path difference is 1cm.

**ThirstyLearner**)Mark scheme says- π radians is the phase difference between the two waves at A – the path difference between the two waves is (8cm – 5cm) = 3cm = 1.5λ. This results in the waves arriving in antiphase (π radians out of phase) which also occurs when the path difference is 1cm.

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(Original post by

Wait...(8cm-5cm?) how did that happen tho thought it was 6.5-6.5...

**Spannerin'moi**)Wait...(8cm-5cm?) how did that happen tho thought it was 6.5-6.5...

and 2.5+2.5=5

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#11

Thanks for the sharing the question...helped me revise

Last edited by username2889812; 1 year ago

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#12

**ThirstyLearner**)

Mark scheme says- π radians is the phase difference between the two waves at A – the path difference between the two waves is (8cm – 5cm) = 3cm =

**1.5λ.**This results in the waves arriving in antiphase (π radians out of phase) which also occurs when the

**path difference is 1cm.**

Last edited by username2889812; 1 year ago

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#14

The path difference is (4+4)-(2.5+2.5) = 8 - 5 = 3 cm

3 cm is equal to 1.5 wavelengths

The path difference is 1.5 wavelengths. We have destructive interference and the phase difference is pi.

[1.5 wavelengths corresponds to a phase difference of 3 pi which is equivalent to a phase difference of pi]

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#15

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Yes. We ignore any phase change during reflection [if interested on this : https://physics.stackexchange.com/qu...-on-refraction, ]https://physics.stackexchange.com/questions/32122/phase-shift-of-180-degrees-of-transversal-wave-on-reflection-from-denser-medium]

The path difference is (4+4)-(2.5+2.5) = 8 - 5 = 3 cm

3 cm is equal to 1.5 wavelengths

The path difference is 1.5 wavelengths. We have destructive interference and the phase difference is pi.

[1.5 wavelengths corresponds to a phase difference of 3 pi which is equivalent to a phase difference of pi]

**BobbJo**)Yes. We ignore any phase change during reflection [if interested on this : https://physics.stackexchange.com/qu...-on-refraction, ]https://physics.stackexchange.com/questions/32122/phase-shift-of-180-degrees-of-transversal-wave-on-reflection-from-denser-medium]

The path difference is (4+4)-(2.5+2.5) = 8 - 5 = 3 cm

3 cm is equal to 1.5 wavelengths

The path difference is 1.5 wavelengths. We have destructive interference and the phase difference is pi.

[1.5 wavelengths corresponds to a phase difference of 3 pi which is equivalent to a phase difference of pi]

I got it till the destructive interference part....

Is it a rule of thumb that all destructive interferences should have a phase difference of π?

And also how do I plug in values into this equation?

Φ = (2π/λ) * x ?

I thought this was to be done...

(2π/λ) * 1.5λ but it results in 3 π

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#16

(Original post by

for extra links and yer patience

And also how do I plug in values into this equation?

Φ = (2π/λ) * x ?

I thought this was to be done...

(2π/λ) * 1.5λ but it results in 3 π

**Spannerin'moi**)for extra links and yer patience

And also how do I plug in values into this equation?

Φ = (2π/λ) * x ?

I thought this was to be done...

(2π/λ) * 1.5λ but it results in 3 π

For the equation, plug in values of λ and x

Φ = (2π/λ) * x

in this question, λ = 2 cm

hence Φ = (2π/2) x 3 = 3π

or convert x = 3 cm to x = 3/2 λ and replace, thus

Φ = (2π/λ) * 3/2 λ = 3π

I got it till the destructive interference part....

Is it a rule of thumb that all destructive interferences should have a phase difference of π?

Is it a rule of thumb that all destructive interferences should have a phase difference of π?

In fact, the path difference must be half a wavelength, or (n+1/2)λ where n is an integer.

This corresponds to a phase difference of Φ = (2π/λ) * x = (2n+1)π for integer n.

Destructive interference therefore occurs for a phase difference: π, 3π, 5π, 7π

But all of these are equivalent to a phase difference of π

The displacement of a point on a wave varies sinusoidally:

x = x

_{0}sin(ωt+Φ), Φ denotes the phase angle

Look at the following:

https://www.desmos.com/calculator/epjylr2j7v

Notice that the curves are identical for Φ = π, 3π, 5π, ...

This is due to the periodicity of the sine curve. Hence phase difference of 3 pi and 5 pi are equivalent to a phase difference of pi

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#17

(Original post by

[Note: the reflection can be ignored because both will undergo reflection and hence both will have the pi phase change, hence it cancels out]

For the equation, plug in values of λ and x

Φ = (2π/λ) * x

in this question, λ = 2 cm

hence Φ = (2π/2) x 3 = 3π

or convert x = 3 cm to x = 3/2 λ and replace, thus

Φ = (2π/λ) * 3/2 λ = 3π

Yes, all destructive interferences occur when the phase difference is π.

In fact, the path difference must be half a wavelength, or (n+1/2)λ where n is an integer.

This corresponds to a phase difference of Φ = (2π/λ) * x = (2n+1)π for integer n.

Destructive interference therefore occurs for a phase difference: π, 3π, 5π, 7π

But all of these are equivalent to a phase difference of π

The displacement of a point on a wave varies sinusoidally:

x = x

Look at the following:

https://www.desmos.com/calculator/epjylr2j7v

Notice that the curves are identical for Φ = π, 3π, 5π, ...

This is due to the periodicity of the sine curve. Hence phase difference of 3 pi and 5 pi are equivalent to a phase difference of pi

**BobbJo**)[Note: the reflection can be ignored because both will undergo reflection and hence both will have the pi phase change, hence it cancels out]

For the equation, plug in values of λ and x

Φ = (2π/λ) * x

in this question, λ = 2 cm

hence Φ = (2π/2) x 3 = 3π

or convert x = 3 cm to x = 3/2 λ and replace, thus

Φ = (2π/λ) * 3/2 λ = 3π

Yes, all destructive interferences occur when the phase difference is π.

In fact, the path difference must be half a wavelength, or (n+1/2)λ where n is an integer.

This corresponds to a phase difference of Φ = (2π/λ) * x = (2n+1)π for integer n.

Destructive interference therefore occurs for a phase difference: π, 3π, 5π, 7π

But all of these are equivalent to a phase difference of π

The displacement of a point on a wave varies sinusoidally:

x = x

_{0}sin(ωt+Φ), Φ denotes the phase angleLook at the following:

https://www.desmos.com/calculator/epjylr2j7v

Notice that the curves are identical for Φ = π, 3π, 5π, ...

This is due to the periodicity of the sine curve. Hence phase difference of 3 pi and 5 pi are equivalent to a phase difference of pi

a ton for the detailed explanation! I understand better now and hopefully OP

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