Q172 / 180 · COMEDK 2012 · Physics

In an interference pattern the ratio of the intensity of light at the bright fringe to that at the dark fringe is $9: 1$. Then, the ratio of the amplitudes of the two interfering waves

  • A
    $3: 1$
  • Correct answer
    B
    $2: 1$
  • C
    $1: 4$
  • D
    $5: 4$

Solution

Given, $\frac{I_{\max }}{I_{\min }}=9$ Since, intensity $\propto(\text { amplitude })^{2}$ $\therefore \quad \frac{A_{\max }}{A_{\min }}=\frac{a_{1}+a_{2}}{a_{1}-a_{2}}=\sqrt{\frac{9}{1}}=\frac{3}{1}$ $\Rightarrow \quad a_{1}+a_{2}=3 a_{1}-3 a_{2}$ $\Rightarrow \quad 2 a_{1}=4 a_{2}$ $\therefore \quad a_{1}: a_{2}=2: 1$

Hence, the correct answer is option: 'B'.

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