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Hi there,

I am unsure of how to calculate the path difference in questions relating to constructive and destructive interference.

I understand that when two waves are exactly in phase with each other, constructive interference occurs, so I labeled the intersecting points as areas where constructive interference occurs. But the answer scheme states that at the points where the waves are in phase, the path difference is $nλ$, while at the points where destructive interference occurs, the path difference is $(n − 1/2)\lambda$.

Where do these equations come from?

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Re. "Where do these equations come from?"

Note that $sin(\theta + n\lambda)$ = $sin(\theta)$ so a phase shift of $n\lambda$ leaves the two waves in phase so they reinforce.

Whereas $sin(\theta + (n-1/2)\lambda)$ = $-sin(\theta)$ so a phase shift of $(n-1/2)\lambda$ leaves the two waves 180 degrees out of phase so the cancel.

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You probably know that the total wave in space is simply the sum of the waves from points $A$ and $B$. What matters for the strength of a signal is the intensity of a wave, as its squared modulus. Then, the maximum of constructive interference (point X) occurs when two peaks - represented by red circles - or two troughs - represented by green circles - meet. On the other hand, the minimum (point Y) will be where a peak meets a trough.

As you can see here, in a somewhat different problem: at $x=0$, both waves have values $w_1(0)=w_2(0)=1$, so they interfere constructively, while at $x=\pi$ we have $w_1(\pi)=-w_2(\pi)=-1$, so the intensity is 0 and the interference is destructive. interference

The other part of your question is about the path differences. In your specific case, notice that the peaks happen at integer multiples $n\lambda$ of the wavelength, and the troughs happen at semi-integers $(n-1/2)\lambda$ away from the source. If a maximum is at the intersection of the red circles, the path difference is $$ d = n_A\lambda - n_B\lambda = (n_A-n_B)\lambda=n\lambda,\ \text{ with }n\text{ integer}, $$ and if it is at a intersection of green circles, $$ d = \left(n_A-\frac{1}{2}\right)\lambda - \left(n_B-\frac{1}{2}\right)\lambda = \left(n_A-n_B\right)\lambda=n\lambda,\ \text{ with }n\text{ integer}. $$

If it is at a minimum, it is at a point where a red circle meets a green circle, so we have a situation like $$ d = \left(n_A-\frac{1}{2}\right)\lambda -n_B\lambda = \left(n_A-n_B-\frac{1}{2}\right)\lambda=\left(n-\frac{1}{2}\right)\lambda,\ \text{ with }n\text{ integer}. $$ Notice that the $-1/2$ might be on the other parentheses, it doesn't change the result.

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Let say you have two electromagnetic wave

$$y_{1}=A\sin(\omega t-kx_{1}),\,y_{2}=A\sin(\omega t-kx_{2}),$$

Now phase difference denoted, by variable $h$, so we get $$kx_{1}-kx_{2}=h\tag{1}$$

Where $k$ denote wave number, because we measure the distance traveled by wave in terms of wavelength, so we get the wave number.

Where $(x_{1}-x_{2})$ is the path difference. Two waves interfere like a vector did.

So when the phase difference is integral of $2n(\pi)$,then wave will show constructive interference, if it $(n+1/2)(\pi)$ then they show destructive interference, after this value in equation 1, you will get path difference between waves. As a function of $n$.

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  • $\begingroup$ Hi man let's leave the rift between us and be friends.I just recommend you that please edit the answer please replace A be E as the amplitudes here are electric fields.Please don't accept my edit.I did something wrong.🤝🤝 $\endgroup$ – RunMachine_Kohli Oct 26 '19 at 10:19
  • $\begingroup$ I apologize for everything, can you come in this chat room for once. chat.stackexchange.com/rooms/100303/physics-for-indian-exams $\endgroup$ – Yuvraj Singh... Oct 26 '19 at 10:38
  • $\begingroup$ Yes I am Online right now $\endgroup$ – RunMachine_Kohli Oct 26 '19 at 10:40
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They come from the realisation that a sine or cosine function changes sign twice every wavelength.  

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  • $\begingroup$ Not wrong just understatement. Added twice 🤓 $\endgroup$ – my2cts Oct 26 '19 at 20:55

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