Q127 / 160 · WBJEE 2010 · Physics

The light beams of intensities in the ratio of $9: 1$ are allowed to interfere. What will be the ratio of the intensities of maxima and minima?

  • A
    $3: 1$
  • Correct answer
    B
    $4: 1$
  • C
    $25: 9$
  • D
    $81: 1$

Solution

Hints: $\frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{3}{1}$ $$ \frac{I_{\max }}{I_{\min }}=\frac{16}{4}=\frac{4}{1} $$

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

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