MCQ
Light waves producing interference have their amplitudes in the ratio $3 : 2$. The intensity ratio of maximum and minimum of interference fringes is
- A$36:1$
- B$9:4$
- ✓$25:1$
- D$6:4$
✓
Answer
Correct option: C.
$25:1$
c
(c)$\frac{{{I_{\max }}}}{{{I_{\min }}}} = {\left( {\frac{{\frac{{{a_1}}}{{{a_2}}} + 1}}{{\frac{{{a_1}}}{{{a_2}}} - 1}}} \right)^2} = \frac{{25}}{1}$
(c)$\frac{{{I_{\max }}}}{{{I_{\min }}}} = {\left( {\frac{{\frac{{{a_1}}}{{{a_2}}} + 1}}{{\frac{{{a_1}}}{{{a_2}}} - 1}}} \right)^2} = \frac{{25}}{1}$
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