To find the resultant intensity

• PSN03
In summary, the conversation discussed calculating the path differences between three points and finding their phase differences using the equation Cosx=2π/λ* (path difference). It was mentioned that the method of adding the intensities of three sources would not work and instead, the amplitudes should be added and squared. One participant suggested starting with a convenient choice of phase, and using this method, the conversation ended with a plan to continue solving the problem the following day.
PSN03
Homework Statement
Consider the interference at P between waves emanating from three coherent sources in same phase located at S1, S2 and S3. If intensity due to each source is I0 = 12 W/m2 at P and d2/2D =λ/3 calculate the resultant intensity at P.
Relevant Equations
I= I1+I2+ 2√I1*I2cosx
Where I=net intensity
x=phase difference
So I thought of calculating the path differences between all the 3 points by taking them in pairs of 2.
S1 and S2
S2 and S3
S3 and S1
I got the path differences as λ/3 , λ and 2λ/3
I can now find their phase differences using the equation
Cosx=2π/λ* (path difference)
Then I can apply the intensity equation to find 3 intensities obtained from 3 different pairs.
Is my logic correct? If yes then can I directly take the summation of these 3 intensities now?

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You only need to find two phases, not three because it is the phase difference that counts. I would write the amplitudes for each wave,
##A_1=\sqrt{I_0}\cos(kD)##
##A_2=\sqrt{I_0}\cos(kx_2)##
##A_3=\sqrt{I_0}\cos(kx_3)##
then translate the cosine arguments into phases involving the wavelength, add the amplitudes and finally square. That would be safe.

Last edited:
vela
PSN03 said:
Then I can apply the intensity equation to find 3 intensities obtained from 3 different pairs.
Is my logic correct? If yes then can I directly take the summation of these 3 intensities now?
No. That method wouldn't work. It would be like saying ##(a+b+c)^2 = (a+b)^2 + (b+c)^2 + (a+c)^2##.

When you have coherent sources, intensities don't add. Amplitudes do. You want to do as @kuruman suggests.

PSN03
kuruman said:
You only need to find two phases, not three because it is the phase difference that counts. I would write the amplitudes for each wave,
##A_1=\sqrt{I_0}\cos(kD)##
##A_2=\sqrt{I_0}\cos(kx_2)##
##A_3=\sqrt{I_0}\cos(kx_3)##
then translate the cosine arguments into phases involving the wavelength, add the amplitudes and finally square. That would be safe.
Sorry but how is it 2 only? Shouldn't it be 3?
Are you trying to say that each source will produce a particular amplitude at the point P?
I clearly got your adding the amplitude and and then squaring it.
But I didn't get the upper one.

PSN03 said:
Sorry but how is it 2 only? Shouldn't it be 3?
Are you trying to say that each source will produce a particular amplitude at the point P?
I clearly got your adding the amplitude and and then squaring it.
But I didn't get the upper one.
We cannot help you unless you show us exactly what you did. To explain what I meant: more correctly the amplitudes are time-dependent and one should write
##A_1(t)=\sqrt{I_0}\cos(kD-\omega t)##
##A_2(t)=\sqrt{I_0}\cos(kx_2-\omega t)##
##A_3(t)=\sqrt{I_0}\cos(kx_3-\omega t)##
Now suppose I pick a specific time ##t_s## such that ##(kD-\omega t_s)=256~\pi## or any other even multiple of ##\pi##. What are ##A_1(t_s)##, ##A_2(t_s)## and ##A_3(t_s)##? Add 'em up and square.

PSN03
kuruman said:
We cannot help you unless you show us exactly what you did. To explain what I meant: more correctly the amplitudes are time-dependent and one should write
##A_1(t)=\sqrt{I_0}\cos(kD-\omega t)##
##A_2(t)=\sqrt{I_0}\cos(kx_2-\omega t)##
##A_3(t)=\sqrt{I_0}\cos(kx_3-\omega t)##
Now suppose I pick a specific time ##t_s## such that ##(kD-\omega t_s)=256~\pi## or any other even multiple of ##\pi##. What are ##A_1(t_s)##, ##A_2(t_s)## and ##A_3(t_s)##? Add 'em up and square.
Can I upload my work tomorrow as its very late at my place and I need to sleep

PSN03 said:
Can I upload my work tomorrow as its very late at my place and I need to sleep
Of course. Take your time and if I am not available, someone else will be.

PSN03
ohk,
so my strategy was to find the interference pattern due to the 3 sources by taking them in the pairs of 2
Δ ##x_1=S_1P-S_2P##
=D-(##D^2+d^2##)^(1/2)
=D-(D(1-##d^2/2D^2##)) (by binomial expansion)
=##d^2/2D##
= λ/3 (from provided relation)

Similarly
Δ ##x_2=S_2P-S_2P##
=##9d^2/4D##
=3λ/4
and
Δ ##x_3=S_1P-S_3P##
=##2d^2/D##
=4λ/3

Now we can calculate the phase difference using the formula
ΔΦ=2 π /λ*Δ x

Through this phase difference we can calculate the resultant intensity due to the 2 sources.
And as we know I is proportional to ##A^2## we can hence find A and perform its summation.
This was my working. Can you please point the mistake

You are not following my suggestion. Don't worry about path length differences. Start with ##\cos(kD-\omega t_s)=1##, a convenient choice of phase. It implies that the phase of ##A_1## is a multiple of 2##\pi##. This makes ##A_1=\sqrt{I_0}##.

Now find ##A_2##. You have found that ##x_2=D+\dfrac{d^2}{2D}=D+\dfrac{\lambda}{3}.## What is ##\cos(kx_2-\omega t_s)?## Hint: ##k=\dfrac{2\pi}{\lambda}## and you already know that ##(kD-\omega t_s)## is a multiple of ##2\pi##.

Now it's my turn to get some sleep.

[/QUOT
kuruman said:
You are not following my suggestion. Don't worry about path length differences. Start with ##\cos(kD-\omega t_s)=1##, a convenient choice of phase. It implies that the phase of ##A_1## is a multiple of 2##\pi##. This makes ##A_1=\sqrt{I_0}##.

Now find ##A_2##. You have found that ##x_2=D+\dfrac{d^2}{2D}=D+\dfrac{\lambda}{3}.## What is ##\cos(kx_2-\omega t_s)?## Hint: ##k=\dfrac{2\pi}{\lambda}## and you already know that ##(kD-\omega t_s)## is a multiple of ##2\pi##.

Now it's my turn to get some sleep.
Actually I am not aware of the method you are trying to apply. I have never learned this so I can't apply.
And shouldn't ##x_2##=D-d²/2D?

By the way I have a solution to this problem and it is attached below.
I was trying to solve this problem by applying the idea from this solution though I couldn't do any good.
See you later then, good night.

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kuruman said:
You are not following my suggestion. Don't worry about path length differences. Start with ##\cos(kD-\omega t_s)=1##, a convenient choice of phase. It implies that the phase of ##A_1## is a multiple of 2##\pi##. This makes ##A_1=\sqrt{I_0}##.

Now find ##A_2##. You have found that ##x_2=D+\dfrac{d^2}{2D}=D+\dfrac{\lambda}{3}.## What is ##\cos(kx_2-\omega t_s)?## Hint: ##k=\dfrac{2\pi}{\lambda}## and you already know that ##(kD-\omega t_s)## is a multiple of ##2\pi##.

Now it's my turn to get some sleep.
In light of the above hints you dropped I can say
kD=wt+2πn

Hence I can say
Cos(k##x_2##-wt)
=Cos(k(D-d²/2D)-wt)
=Cos(-kd²/2D)
=Cos(-2π/λ*λ/3)
=Cos(2π/3)
=-1/2

Now according to the formula you have given my
##A_2##=√I*(-1/2)
=-√I/2

Am I doing right?
I don't know why we hould apply this, I am just following your instructions
Can you please teall me why are we doing this cause I have never done any question with this method.

kuruman said:
You are not following my suggestion. Don't worry about path length differences. Start with ##\cos(kD-\omega t_s)=1##, a convenient choice of phase. It implies that the phase of ##A_1## is a multiple of 2##\pi##. This makes ##A_1=\sqrt{I_0}##.

Now find ##A_2##. You have found that ##x_2=D+\dfrac{d^2}{2D}=D+\dfrac{\lambda}{3}.## What is ##\cos(kx_2-\omega t_s)?## Hint: ##k=\dfrac{2\pi}{\lambda}## and you already know that ##(kD-\omega t_s)## is a multiple of ##2\pi##.

Now it's my turn to get some sleep.
@kuruman I saw some videos online and I got the message you are trying to convey. You are going to find the resultant amplitude cause by 3 sources through the vector addition of individual amplitudes (or by their phasor diagrams). I now under stand what you were doing. If my interpretation was correct you are finding their phase differences by considering the first source as main source with no phase difference.

Now my doubt is if we take any two sources, they will produce an interference pattern on point P which will have some intensity I and some amplitude A.
Since there are 3 sources hence 3 such pairs are possible. So why can't we calculate the intensity due to these 3 interference patterns and find their amplitude and then perform the summation?

PSN03 said:
Since there are 3 sources hence 3 such pairs are possible. So why can't we calculate the intensity due to these 3 interference patterns and find their amplitude and then perform the summation?
That's not how it works and @vela explained why in post #3. The intensity at a given point is the amplitude squared of whatever electric field exists at that point. If you have a single source it is the square of that single source and you don't have to add anything. If you have three sources, first you add as vectors the fields from all three to get the total field and then square.

My method is exactly that. Calculate the amplitudes, then add them then square the sum. The method of path length differences that you are trying to apply works well when you have either constructive (maximum intensity) or destructive (zero intensity) interference, but what is the intensity if the path length difference between two sources is (3/4}λ? You get "in-between" interference and to find the intensity you need to add the amplitudes and square the result. An example of in-between interference is this problem with S3 missing. The total amplitude from S1 and S2 is ##A_{tot}=\sqrt{I_0}+(-\frac{1}{2})\sqrt{I_0}=(\frac{1}{2})\sqrt{I_0}## in which case the intensity is ##I_{tot}=\frac{1}{4}I_0##.

You have made good progress and you calculation in post #11 is correct. Repeat to find ##\cos(kx_3-\omega t_s)## and then add the three amplitudes.

PSN03
kuruman said:
That's not how it works and @vela explained why in post #3. The intensity at a given point is the amplitude squared of whatever electric field exists at that point. If you have a single source it is the square of that single source and you don't have to add anything. If you have three sources, first you add as vectors the fields from all three to get the total field and then square.

My method is exactly that. Calculate the amplitudes, then add them then square the sum. The method of path length differences that you are trying to apply works well when you have either constructive (maximum intensity) or destructive (zero intensity) interference, but what is the intensity if the path length difference between two sources is (3/4}λ? You get "in-between" interference and to find the intensity you need to add the amplitudes and square the result. An example of in-between interference is this problem with S3 missing. The total amplitude from S1 and S2 is ##A_{tot}=\sqrt{I_0}+(-\frac{1}{2})\sqrt{I_0}=(\frac{1}{2})\sqrt{I_0}## in which case the intensity is ##I_{tot}=\frac{1}{4}I_0##.

You have made good progress and you calculation in post #11 is correct. Repeat to find ##\cos(kx_3-\omega t_s)## and then add the three amplitudes.
Really sorry sir...I got my mistake
I am sooo stupid to do such kind of mistake
I have got all the poin you are telling me. Thanks a lot sir
One last question...is this relation A=√Icos(kx-wt) valid for all cases?

PSN03 said:
Really sorry sir...I got my mistake
I am sooo stupid to do such kind of mistake
I have got all the poin you are telling me. Thanks a lot sir
One last question...is this relation A=√Icos(kx-wt) valid for all cases?
It is a simplified solution to the electromagnetic wave equation. More correctly one writes the electric field vector for a plane electromagnetic wave traveling in the +x-direction as
$$\vec E(x,t)=\vec E_0\sin(kx-\omega t+\phi).$$Here, ##\vec E_0## is a vector in a plane perpendicular to the x-axis. Note that the phase angle ##\phi## can be chosen so that the field has the value you want at position ##x=0## and time ##t=0##. If you choose ##\phi=\pi/2##, you get the sine to become a cosine, which is what you have here.

PSN03
kuruman said:
It is a simplified solution to the electromagnetic wave equation. More correctly one writes the electric field vector for a plane electromagnetic wave traveling in the +x-direction as
$$\vec E(x,t)=\vec E_0\sin(kx-\omega t+\phi).$$Here, ##\vec E_0## is a vector in a plane perpendicular to the x-axis. Note that the phase angle ##\phi## can be chosen so that the field has the value you want at position ##x=0## and time ##t=0##. If you choose ##\phi=\pi/2##, you get the sine to become a cosine, which is what you have here.
Thank you for your time, patience and work. Means a lot to me.
Good day and stay safe

1. What is the formula for finding the resultant intensity?

The formula for finding the resultant intensity is I = I1 + I2 + 2√(I1I2)cosθ, where I1 and I2 are the individual intensities and θ is the angle between them.

2. How do I determine the direction of the resultant intensity?

The direction of the resultant intensity can be determined by using the formula θ = tan^-1(I2 sinθ / (I1 + I2 cosθ)), where θ is the angle between the resultant intensity and I1.

3. Can the resultant intensity ever be greater than the sum of the individual intensities?

Yes, the resultant intensity can be greater than the sum of the individual intensities if the angle between them is less than 90 degrees. This is known as constructive interference.

4. How does the angle between the individual intensities affect the resultant intensity?

The angle between the individual intensities determines whether the resultant intensity will be greater or less than the sum of the individual intensities. If the angle is less than 90 degrees, the resultant intensity will be greater due to constructive interference. If the angle is greater than 90 degrees, the resultant intensity will be less due to destructive interference.

5. Can the resultant intensity ever be negative?

No, the resultant intensity cannot be negative. Intensity is a measure of the amount of energy per unit area, and it is always a positive value. If the calculated resultant intensity is negative, it means there is an error in the calculation.

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