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Two coherent monochromatic light beams of intensities \mathrm{I} and \mathrm{4I} are superimposed. The maximum and minimum possible intensities in the resulting beam are:

Option: 1

5 \mathrm{I}$ and $ \mathrm{I}


Option: 2

5 \mathrm{I}$ and $3 \mathrm{I}


Option: 3

9 \mathrm{I}$ and $\mathrm{I}


Option: 4

9 \mathrm{I}$ and $3 \mathrm{I}


Answers (1)

best_answer

\quad \text { Intensity } \alpha(\text { Amplitude) })^{2}
\Rightarrow \quad I \propto A^{2}

When two waves (beams) of amplitude A_{1} and A_{2} superimpose, at maxima and minima, the amplitude of the resulting wave are \left(A_{1}+\right.$ $\left.A_{2}\right)$ and $\left(A_{1}-A_{2}\right) respectively. If the maximum and minimum possible intensities are I_{\max }$ and $I_{\min } respectively, then

I_{\max } \alpha\left(A_{1}+A_{2}\right)^{2}
\text { And } \quad I_{\min } \alpha\left(A_{1}-A_{2}\right)^{2}

\Rightarrow \quad \frac{I_{\max }}{I_{\min }}=\left(\frac{A_{1}+A_{2}}{A_{1}-A_{2}}\right)^{2}=\left\{\frac{\frac{A_{1}}{A_{2}}+1}{\frac{A_{1}}{A_{2}}-1}\right\}^{2} \text { where } \frac{A_{1}}{A_{2}}=\frac{\sqrt{I}}{\sqrt{4 I}}=\frac{1}{2}
\Rightarrow \quad \frac{I_{\max }}{I_{\min }}=\frac{9}{1} \Rightarrow I_{\max }=9 I, I_{\min }=I

Posted by

Irshad Anwar

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