Optics Ques 145
- Two coherent monochromatic light beams of intensities $I$ and $4 $ $I$ are superposed. The maximum and minimum possible intensities in the resulting beam are
$(1988,1 M)$
(a) $5 $ $ I$ and $ I$
(b) $5 $ $ I$ and $3 $ $ I$
(c) $9 $ $ I$ and $ I$
(d) $9 $ $ I$ and $3 $ $ I$
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Answer:
Correct Answer: 145.(c)
Solution:
Formula:
For Destructive interference :
- $I _{\text {max }}=\left(\sqrt{I _1}+\sqrt{I _2}\right)^{2}=(\sqrt{4 I}+\sqrt{I})^{2}=9 I$
$I _{\text {min }}=\left(\sqrt{I _1}-\sqrt{I _2}\right)^{2}=(\sqrt{4 I}-\sqrt{I})^{2}=I$
BETA
