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Two silts in young’s experiment have width in the ratio 1:25. The ratio of intensity at maxima and minima in the interference pattern \frac{I_{max}}{I_{min}} is:

Option: 1

\frac{9}{4}


Option: 2

\frac{3}{2}


Option: 3

\frac{121}{49}


Option: 4

\frac{5}{1}


Answers (1)

best_answer

As we learn

Maximum amplitude & Intensity -

When \theta = 0,2\pi ---2n\pi
 

- wherein

A_{max }= A_{1}+A_{2}

I_{max }= \left ( \sqrt{I_{1}}+\sqrt{I_{2}} \right )^{2}

 

 \frac{I_{1}}{I_{2}}=\frac{1}{25}                  OR              \frac{I_{2}}{I_{1}}=\frac{25}{1}

\frac{I_{max}}{I_{min}}=\left ( \frac{\sqrt{I_{2}}+\sqrt{I_{1}}}{\sqrt{I_{2}}-\sqrt{I_{1}}} \right )^{2}        \frac{I_{max}}{I_{min}}=\left ( \frac{\sqrt{I_{2}}+\sqrt{I_{1}}}{\sqrt{I_{2}}-\sqrt{I_{1}}} \right )^{2}= \left ( \frac{\sqrt{\frac{I_{2}}{I_{1}}}+1}{\sqrt{\frac{I_{2}}{I_{1}}}-1} \right )^{2}

\frac{I_{max}}{I_{min}}=\left ( \frac{5+1}{5-1} \right )^{2}=\frac{9}{4}

 

Posted by

Anam Khan

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