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Suppose I have two waves: $Y_1= a_1e^{i(wt-kx1)}$ and $Y_2= a_2e^{i(wt-kx2)}$ I know by superposition $Y= Y_1+Y_2$ and intensity $(I) = |Y|² $ But how can I solve it. It seems hard for me to find the intensity.

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$$Y_1= a_1e^{i(wt-kx1)},Y_2= a_2e^{i(wt-kx2)}$$ $$Y^*_1= a_1e^{-i(wt-kx1)},Y^*_2= a_2e^{-i(wt-kx2)}$$ $$Y=Y_1 + Y_2$$ $$|Y|²=YY^*=Y_1(Y_1^*+Y_2^*)+Y_2(Y_1^*+Y_2^*)$$ $$Y_1Y_1^*=a_1^2$$ $$Y_2Y_2^*=a_2^2$$ $$Y_1Y_2^*=a_1a_2e^{-ik(x_2-x_1)}$$ $$Y_2Y^*_1=a_1a_2e^{ik(x_2-x_1)}$$

$$YY^*=a_1^2+a_2^2+a_1a_2(e^{-ik(x_2-x_1)}+e^{ik(x_2-x_1)})$$ $$|Y|²=a_1^2+a_2^2+2a_1a_2cos[k(x_2-x_1)]$$

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  • $\begingroup$ Thanks a lot. Finally I understand it totally. Thanks to everyone. $\endgroup$ – Saikat Sengupta Apr 9 at 4:56
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Summing

$$Y_1 = a_1 \, e^{i(\omega_1 t - k_1 x)} = a_1 \cos(\omega_1 t - k_1 x) + i \, a_2 \sin(\omega_1 t - k_1 x)$$

and

$$Y_2 = a_2 \, e^{i(\omega_2 t - k_2 x)} = a_2 \cos(\omega_2 t - k_2 x) + i \, a_2 \sin(\omega_2 t - k_2 x)$$

(which is what I assume you meant to do) gives

$$Y_1 + Y_2 = \Big( a_1 \cos(\omega_1 t - k_1 x) + a_2 \cos(\omega_2 t - k_2 x) \Big) + i \Big( a_2 \sin(\omega_1 t - k_1 x) + a_2 \sin(\omega_2 t - k_2 x) \Big) $$

and

$$|Y_1 + Y_2|^2 = \Big( a_1 \cos(\omega_1 t - k_1 x) + a_2 \cos(\omega_2 t - k_2 x) \Big)^2 + \Big( a_2 \sin(\omega_1 t - k_1 x) + a_2 \sin(\omega_2 t - k_2 x) \Big)^2 = a_1^2 \cos^2(\omega_1 t - k_1 x) + 2 a_1 a_2 \cos(\omega_1 t - k_1 x) \cos(\omega_2 t - k_2 x) + a_2^2 \cos^2(\omega_2 t - k_2 x) + a_1^2 \sin^2(\omega_1 t - k_1 x) + 2 a_1 a_2 \sin(\omega_1 t - k_1 x) \sin(\omega_2 t - k_2 x) + a_2^2 \sin^2(\omega_2 t - k_2 x)$$

which works out to

$$ |Y_1 + Y_2|^2 = a_1^2 + a_2^2 + 2 a_1 a_2 \Big( \cos(\omega_1 t - k_1 x) \cos(\omega_2 t - k_2 x) + \sin(\omega_1 t - k_1 x) \sin(\omega_2 t - k_2 x) \Big) $$

and using the product formulas for sine and cosine to

$$ a_1^2 + a_2^2 + 2 a_1 a_2 \cos\Big( (\omega_1 - \omega_2)t - (k_1 - k_2)x \Big) $$

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  • $\begingroup$ Thanks a lot. But if I want to do it as I = (Y*Y) like the complex conjugate process then how to solve it? $\endgroup$ – Saikat Sengupta Apr 7 at 15:46
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I am new to the site and I need to write as an answer what I would like to write as a comment.

The comment is: what is it that you are able to do? You already know that

$I = |Y|^2 = YY^*,$

that is you can calculate the modulus square of a complex number by multiplying it times its complex conjugate. The next step is substituting into the expression above the sum $Y_1 + Y_2$ for $Y$; are you able to progress further from this point? That is, where is it that you get stuck?

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