Optics Ques 2
2 Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16 . The intensity of the waves are in the ratio
(2019 Main, 9 Jan I)
(a) $16: 9$
(b) $5: 3$
(c) $25: 9$
(d) $4: 1$
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Answer:
Correct Answer: 2.( c )
Solution:
- Let the intensity of two coherent sources be $I_1$ and $I_2$, respectively. It is given that,
$ \frac{\text { maximum intensity }}{\text { minimum intensity }}=\frac{I_{\max }}{I_{\min }}=\frac{16}{1} $
Since, we know and $ \begin{aligned} & I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2 \\ & I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2 \end{aligned} $
$\therefore$ We can write,
$ \begin{aligned} & \frac{I_{\max }}{I_{\text {min }}}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2}=16 \\ & \Rightarrow \quad \frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}=\frac{\sqrt{16}}{1}=\frac{4}{1} \\ & \sqrt{I_1}+\sqrt{I_2}=4 \sqrt{I_1}-4 \sqrt{I_2} \\ & 5 \sqrt{I_2}=3 \sqrt{I_1} \Rightarrow \frac{\sqrt{I_1}}{\sqrt{I_2}}=\frac{5}{3} \\ & \end{aligned} $
$ \Rightarrow $
Squaring on both the sides, we get
$ \frac{I_1}{I_2}=\frac{25}{9} $
BETA
