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Monochromatic light waves of amplitudes $E_{10}$ and $E_{20}$ and a constant phase difference $\varphi$ produce an interference pattern. State an expression for the resultant amplitude at a point in the pattern. Hence, deduce the conditions for
$(i)$ constructive interference with maximum intensity
$(ii)$ destructive interference with minimum intensity. Also show that the ratio of the maximum and minimum intensities is
$\frac{I_{\max }}{I_{\min }}=\left(\frac{E_{10}+E_{50}}{E_{10}-E_{50}}\right)^2$

Vidyadip · Accepted Answer

Consider a two-source interference pattern produced by monochromatic light waves of angular frequency $\omega$, wavelength $\lambda$ amplitudes $E_{10}$ and $E_{20}$ and a constant phase difference $\varphi$.Let the individual electric fields along the same line due to the waves at some point in the interference pattern be
$E_1=E_{10} \sin \omega t$ and $E_2=E_{20} \sin (\omega t+\varphi)$
By the principle of superposition of waves, the resultant electric field (displacement) at that point is the algebraic sum
$E=E_1+E_2=E_{10} \sin \omega t-E_{20} \sin (e t+\phi)$
$E_{10} \sin \omega t+E_{20} \sin \cot \cos \phi+E_{20} \cos \cot \sin \phi$
$-\left(E_{10}+E_{20} C 00 \phi\right) \sin \omega t+E_{20} \sin \phi \cos \omega t$
$\text { Let } E_{10}+E_{70} \cos \phi-R \cos \theta \text { and }$
$E_{20} \sin \phi-R \sin \theta$
$\therefore E=R \cos \theta \sin \omega t-R \sin \theta \cos \omega t$
$=R(\sin \omega i \cos \theta+\cos \omega i \sin \theta)$
$= R \sin (\omega f +\theta)$
The resultant amplitude is,
$|R|=\sqrt{E_{10}^2+E_{20}^2+2 E_{10} \tilde{E}_{20} \cos \phi}$
Since, the intensity of a wave is proportional to the square of its amplitude, the resultant intensity at $P$,
$\left.|\propto| R\right|^2$
$\therefore l \propto E_{10}^2+E_{20}^2+2 \Gamma_{10} \Gamma_{20} 005 $
Thus, the intensity depends on $\cos \varphi$.
The condition for constructive interference with maximum intensity is $\cos \varphi$ is maximum, equal to $1$ , i.e., $\varphi=2 n \pi(n=0,1,2,3 \ldots)$ (3)
The condition for destructive interference with minimum intensity is $\cos \varphi$ is minimum equal to
$-1, \text { i.e., }$
$\phi=(2 m-1) m \quad(m=1,2,3 \ldots)$
At points where $R$ and $J$ are maximum,
$R_{\min }=E_{10}+E_{20}$
$\text { and } I_{\min } \propto E_{10}^2+E_{20}^2+2 E_{10} E_{20}$
$\therefore I_{\min } \propto\left(E_{10}+E_{20}\right)^2$
At points where $K$ and $I$ are minimum,
$R_{\min }-\left|E_{10}-E_{20}\right|$
$\text { and } d_{\min } \text { o. } E_{C_0}+E ?_0-2 E_{20} E_{20}$
$\therefore I_{\min } \propto\left(L_{10}-L_{20}\right)^2$
$\therefore \text { From liqs. (6) and (B), }$
$\frac{t_{\text {mas }}}{l_{\text {ais }}}-\left(\frac{t_{10}+t_{20}}{t_{10}-t_{20}}\right)^2$
Note : We can write the above expression as
$\frac{L_{\text {ses }}}{I_{\text {wit }}}=\left(\frac{1+E_{29} / E_{19}}{1-E_{29} / E_{i 9}}\right)^{ \pm}=\left(\frac{1+r}{1-r}\right)^2 \text {. }$
where $r$ is the amplitude ratio $\frac{E_{30}}{E_{10}}. 1$

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