Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Four identical monochromatic sources A B C D as shown in the Fig.10 7 produce waves of the same wavelength and are coherent Two receiver R1 and R2 are at great but equal distances

Four identical monochromatic sources A, B, C, D, as shown in the (Fig.10.7) produce waves of the same wavelength and are coherent. Two receiver R1 and R2 are at great but equal distances from B.

 

(i) Which of the two receivers picks up the larger signal?
(ii) Which of the two receivers picks up the larger signal when B is turned off?
(iii) Which of the two receivers picks up the larger signal when D is turned off?
(iv) Which of the two receivers can distinguish which of the sources B or D has been turned off?

Answers (1)

 

(i) Let us consider the disturbances at the receiver R1, which is at a distance d from B.
Let the equation of wave at R1, because of A be

y_A=a \cos \omega t                                                                        ......(i)

The path difference of the signal from A with that from B is   \lambda/2  and hence, the phase difference

\Delta \phi = \frac{2\pi }{\lambda }\times \left ( path\;dif\! \! ference \right )=\frac{2\pi }{\lambda }\times \frac{\lambda }{2}

Thus, the wave equation at R1, because of B is

y_B=a \cos \left ( \omega t-\pi \right )= -a \cos \omega t                                ......(ii)

The path difference of the signal from C with that from A is \lambda and hence the phase difference

\Delta \phi = \frac{2\pi }{\lambda }\times \left ( path\;dif\! \! ference \right )=\frac{2\pi }{\lambda }\times \lambda=2 \pi

Thus, the wave equation at R1 because of C is
y_C=a \cos \omega t =a \cos\left ( \omega t-2\pi \right )= a \cos \omega t            ......(iii)

The path difference between the signal from D with that of A is


\Delta _{x_{R_1}}= \sqrt{d^2+\left ( \frac{\lambda }{2} \right )^{2}}-\left (d- \frac{\lambda }{2} \right )=d\left ( 1+\frac{\lambda ^{2}}{4d^{2}} \right )^{1/2}-d+\frac{\lambda }{2}

            =d\left ( 1+\frac{\lambda ^{2}}{8d^{2}} \right )-d.\frac{\lambda }{2}=\frac{\lambda }{2}\left ( \because d>>\lambda \right )

Therefore, phase difference is \pi

y_D=a \cos\left ( \omega t-\pi \right )= -a \cos \omega t                                ......(iii)

The resultant signal picked up at R_1, from all the four sources is the summation of all four waves,

y_{R_1}=y_A+y_B+y_C+y_D

y_{R_1}=a \cos \omega t-a \cos \omega t+a \cos \omega t-a \cos \omega t=0

Thus, the signal picked up at R1 is zero.

Now let us consider the resultant signal received at R2. Let the equation of wave at R2 . Let the equation
of wave at R2, because of B be

y_{B}=a_1 \cos \omega t

The path difference of the signal from from D with that from B is \frac{\lambda }{2} and

hence, the phase difference

\Delta \phi =\frac{2\pi }{\pi }\times \left ( \text {path difference} \right )=\frac{2\pi}{\lambda }\times \frac{\lambda }{2}=\pi

Thus, the wave equation at R_{2}, because of D is

y_{B}=a_{1}\cos \left ( \omega _{t}-\pi \right )=-a_{1}\cos\; \omega t\; \; \; \; \; \; \; \; \; \; \; \; \; ....(ii)

The path difference between signal at A and that at B is

\Delta x_{R_{2}}=\sqrt{\left ( d \right )^{2}+\left ( \frac{\lambda }{2} \right )^{2}}-d=d\left ( 1+\frac{\lambda ^{2}}{4d^{2}} \right )^{\frac{1}{2}}-d\simeq \frac{\lambda ^{2}}{8d^{2}}

As d>>\lambda , therefore this path differences \Delta x_{R_{2}}\rightarrow 0

and phase difference \Delta \phi =\frac{2\pi}{\lambda }\times \left ( \text {path difference} \right )

                                           =\frac{2\pi}{\lambda }\times 0\rightarrow 0\left ( \text {very small} \right )=\phi \left ( \text {say} \right )

Hence, y_{B}=a_{1}\cos \left ( \omega _{t}-\phi \right )

Similarly, y_{B}=a_{1}\cos \left ( \omega _{t}-\phi \right )

The resultant signal picked up at R_{2}, from all the four sources is the summation of all four waves, y_{R_{2}}=y_{A}+y_{B}+y_{C}+y_{D}

y_{R_{2}}=a_{1}\cos \; \omega t-a_{1}\cos \omega t+a_{1}\cos\left ( \omega t-\phi \right )+a_{1}\cos \left ( \omega t-\phi \right )

          = 2a_{1}\cos \left ( \omega t-\phi \right )

\therefore Signal picked up by R_{2} is y_{R_{2}}=2a_{1}\cos \left ( \omega t-\phi \right )

\therefore               \left [ V_{R_{2}} \right ]^{2}=4a_{1}^{2}\cos^{2}\left ( \omega t-\phi \right )\Rightarrow \left \langle I_{R_{2}} \right \rangle=2a_{1}^{2}

Thus, R_{2} picks up the larger signal.

(ii) If B is switched off,

R_{1}, picks up y=a\cos \; \omega t

\therefore                     \left \langle I_{R_{1}} \right \rangle=a^{2}<\cos^{2}\omega t>=\frac{a^{2}}{2}

R_{2} picks up y=a\; \cos\; \omega t

                         \left \langle I_{R_{2}} \right \rangle=a^{2}<\cos^{2}\omega t>=\frac{a^{2}}{2}

Thus , R_{1} and R_{2} pick up the same signal

(iii) If D is switched off.

      R_{1} picks up y=a\; \cos\; \omega t

\therefore                       \left \langle I_{R_{1}} \right \rangle=\frac{1}{2}a^{2}

R_{2} picks up y=3a\; \cos\; \omega t

\therefore                       \left \langle I_{R_{2}} \right \rangle=9a^{2}<\cos^{2}\omega t>=\frac{9a^{2}}{2}

Thus, R_{2} picks up larger signal compared to R_{1}.

(iv) Thus, a signal at R_{1} indicates B has been switched off and an enhanced signal at R_{2} indicates D has been switched off.

 

Posted by

Gurleen Kaur

View full answer

Similar Questions

Latest Question