Two slits in Youngs experiment have widths in the ratio 1 : 25. The ratio of intensity at the maxima and minima in the interference pattern, \frac{{\text{I}_{\text{max}} }} {{\text{I}_{\text{min}} }} is :

  • Option 1)

    \frac{{\text{121}}} {{\text{49}}}

  • Option 2)

    \frac{{\text{49}}} {{\text{121}}}

  • Option 3)

    \frac{\text{4}} {\text{9}}

  • Option 4)

    \frac{\text{9}} {\text{4}}

 

Answers (1)

I _{2}=25 I, since \frac{I_{2}}{I_{1}}=\frac{W_{2}}{W_{1}}=\frac{25}{1}

 

\therefore \frac{I max}{I min}= \left ( \frac{\sqrt{I_{1}}+\sqrt{25I_{1}}}{\sqrt{I_{1}}-\sqrt{25I_{1}}} \right )^{2}=\left ( \frac{6}{4} \right )^{2}

\frac{I max}{I min}=\frac{9}{4}

 


Option 1)

\frac{{\text{121}}} {{\text{49}}}

This option is incorrect.

Option 2)

\frac{{\text{49}}} {{\text{121}}}

This option is incorrect.

Option 3)

\frac{\text{4}} {\text{9}}

This option is incorrect.

Option 4)

\frac{\text{9}} {\text{4}}

This option is correct.

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