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In the figure below, $\text{P}\; \text{and}\; \text{Q}$ are two equally tense convert sources emitting radiation of wavelength 20 m. The seperation b/w $\text{P}\; \text{and}\; \text{Q}$ is 5 m and the phase of $\text{P}$ is ahead of that of $\text{Q}$ by $90^{\circ}C$ . $\text{A},\text {B}$ and $\text{C}$ are 3 distinct point of observation, each equidistant from the mid-point of $\text{PQ}$ The intransition of radiation at $\text{A},\text {B}$ and $\text{C}$ will be in ratio:-

Option: 1 $2:1:0$
Option: 2 $4:1:0$
Option: 3 $0:1:4$
Option: 4 $0:1:2$

Answers (1)

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$P A-Q A=P Q=5 m=\frac{20}{4}=\frac{\lambda}{4}=\Delta x$$

$\phi^{\prime}=\frac{2 \pi}{\lambda} \Delta x=\frac{2 \pi}{\lambda} \cdot \frac{\lambda}{4}=\frac{\pi}{2}$$

$Phase \ of \ Q$ is greater by $\left(90^{\circ}=\frac{\pi}{2}\right)$ than $P$. \\Net phase difference $\phi=\pi / 2-\pi / 2=0$$

$\begin{aligned} I_{net} &=I_{0}+I_{0}+2 \sqrt{I_{0}^{2}} \cos \phi \\ &=2 I_{0}+2 I_{0} \cos \phi \\ &=2 I_{0}(1+\cos \phi) \end{aligned}$

$I_{A}=4 I_0$

At $\phi=90^{\circ}=\pi / 2$$

$$I_{B}=2 I_0$

At C:

$P C-Q C=P Q=-5 m=-\frac{20}{4}=-\frac{\lambda}{4}=\Delta x$$

$\phi^{\prime}=-\pi / 2$$

$I_{A}: I_{B}: I_{C}=4 I_0: 2 I_0: 0=2: 1: 0$

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Deependra Verma

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