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 Two monochromatic light beams of intensity 16 and 9 units are interfering The ratio of intensities of bright and dark parts of the resultant pattern is :

  • Option 1)

    \frac{16}{9}

  • Option 2)

    \frac{4}{3}

  • Option 3)

    \frac{7}{1}

  • Option 4)

    \frac{49}{1}

 

Answers (1)

best_answer

As we have learned

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}

 

 

Minimum amplitude & Intensity -

\theta = \left ( 2n+1 \right )\pi
 

- wherein

A_{min }= A_{1}-A_{2}

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

 

 \frac{I_{max}}{I_{min}}= \left ( \frac{\sqrt I_1+ \sqrt I_2}{\sqrt I_1- \sqrt I_2} \right )^2 = \left ( \frac{4+3}{4-3} \right )^2 = 49 : 1

 

 


Option 1)

\frac{16}{9}

Option 2)

\frac{4}{3}

Option 3)

\frac{7}{1}

Option 4)

\frac{49}{1}

Posted by

Avinash

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