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  3. In the figure below, are two equally tense convert sources emitting radiation of wavelength
# Engineering

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)
D Deependra Verma

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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