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In young's experiment the interfering has amplitudes in the ratio 3:2, and then ratios of amplitudes between bright and dark fringes are:

Option: 1

5:1


Option: 2

9:4


Option: 3

7:1


Option: 4

49:1


Answers (1)

best_answer

As we learn

Resultant amplitude of two wave -

A= \sqrt{{A_{1}}^{2}+{A_{2}}^{2}+2A_{1}A_{2}\cos \theta }
 

- wherein

A_{1}= amplitude of wave 1

A_{2}=  amplitude of wave 2

\theta = phase difference

 

 We have to obtain the ratio 

\frac{A_{max}}{A_{min}}=\frac{A_{1}+A_{2}}{A_{1}-A_{2}}

and also coressponding ratio of intensities 

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

but \frac{A_{1}}{A_{2}}= \frac{3}{2}

By coresspond and divide 

\frac{A_{1}+A_{2}}{A_{1}-A_{2}}=\frac{3+2}{3-2}=5

Hence ,

\frac{A_{max}}{A_{min}}=5                            AND           \frac{I_{max}}{I_{min}}=25

 

Posted by

Pankaj

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