In an interference experiment the ratio of amplitudes of coherent waves is \frac{a_{1}}{a_{2}}=\frac{1}{3}. The ratio of maximum and minimum intensities of fringes will be :

  • Option 1)

    4

  • Option 2)

    18

  • Option 3)

    2

  • Option 4)

    9

 

Answers (1)

\frac{a_{1}}{a_{2}}=\frac{1}{3}

Intensity \; \alpha \; a^{2}

So, \frac{I_{max}}{I_{min}}=\frac{(a_{1}+a_{2})^{2}}{(a_{1}-a_{2})^{2}}=\frac{(\frac{a_{1}}{a_{2}}+1^{2})}{(\frac{a_{1}}{a_{2}}-1)^{2}}

\frac{I_{max}}{I_{min}}=\frac{(1+3)^{2}}{(1-3)^{2}}=\left ( \frac{4}{2} \right )^{2}=4


Option 1)

4

Option 2)

18

Option 3)

2

Option 4)

9

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