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  • Intensities of two  superposed coherent monochromatic light beams are I and 4I.  The maximum and minimum possible intensities in the resulting beam are <s

Intensities of two  superposed coherent monochromatic light beams are I and 4I.  The maximum and minimum possible intensities in the resulting beam are 

Option: 1

 5I and I

 

 


Option: 2

 5I and 3I


Option: 3

 9I and I


Option: 4

 9I and 3I


Answers (1)

best_answer
 

Resultant Intensity of two waves (I)-

Using I \ \ \alpha \ \ A^2

we get  I= I_{1}+I_{2}+2\sqrt{I_{1}I_{2}}\cos \phi 

Resultant intensity at the point of observation will be maximum

   \begin{aligned} \ \ i.e \ \ I_{\max } &=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}} \\ I_{\max } &=(\sqrt{I_{1}}+\sqrt{I_{2}})^{2} \\ \text { If } \quad I_{1} &=I_{2}=I_{0} \Rightarrow I_{\max }=4 I_{0} \end{aligned}

Resultant intensity at the point of observation will be minimum

    \begin{array}{c}{I_{\min }=I_{1}+I_{2}-2 \sqrt{I_{1} I_{2}}} \\ {I_{\min }=(\sqrt{I_{1}}-\sqrt{I_{2}})^{2}} \\ {\text { If } I_{1}=I_{2}=I_{0} \Rightarrow I_{\min }=0}\end{array}

-

\begin{array}{c}{I_{\max }=(\sqrt{I_{1}}+\sqrt{I_{2}})^{2}=(\sqrt{I}+\sqrt{4 I})^{2}=9 I} \\ {I_{\min }=(\sqrt{I_{1}}-\sqrt{I_{2}})^{2}=(\sqrt{I}-\sqrt{4 I})^{2}=I}\end{array}

Hence, the correct option is (3)

Posted by

Irshad Anwar

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