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Wave Optics Physics Ncert Solutions Class 12th Physics Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Ten

Four identical monochromatic sources A, B, C, D as shown in the (figure)

produce waves of the same wavelength λ 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?

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Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation-consider the disturbance at the receiver R₁ which is at a distance d from B

Y_A= acos(wt) and path difference is $\frac{λ}{2}$ hence phase difference is $π$ .

Thus the wave R₁ because of B

Y_B= acos(wt- $π$ )= - acoswt here path difference is $λ$ and hence phase difference is $2 π$

Thus R₁ because of C

Y_c= acos(wt-2 $π$ )= acoswt

(i)let the signal picked up at R₂ from B be Y_B= a₁cos(wt)

The path difference between signal at D and that B is $\frac{λ}{2}$

Y_D= -a₁cos(wt)

The path difference between signal at A and that atB is

$\sqrt[]{d^{2} + {(\frac{λ}{2})}^{2}}$ -d = d( ${1 + \frac{λ^{2}}{4 D^{2}})}^{1 / 2}$ -d = $\frac{λ^{2}}{8 d^{2}}$

$a s d ? λ$ therefore path difference os 0

$p h a s e d i f f e r e n c e = \frac{2 π}{γ} \frac{λ^{2}}{8 d^{2}}$

_{$Y$
A}=a₁co

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This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation-consider the disturbance at the receiver R₁ which is at a distance d from B

Y_A= acos(wt) and path difference is $\frac{λ}{2}$ hence phase difference is $π$ .

Thus the wave R₁ because of B

Y_B= acos(wt- $π$ )= - acoswt here path difference is $λ$ and hence phase difference is $2 π$

Thus R₁ because of C

Y_c= acos(wt-2 $π$ )= acoswt

(i)let the signal picked up at R₂ from B be Y_B= a₁cos(wt)

The path difference between signal at D and that B is $\frac{λ}{2}$

Y_D= -a₁cos(wt)

The path difference between signal at A and that atB is

$\sqrt[]{d^{2} + {(\frac{λ}{2})}^{2}}$ -d = d( ${1 + \frac{λ^{2}}{4 D^{2}})}^{1 / 2}$ -d = $\frac{λ^{2}}{8 d^{2}}$

$a s d ? λ$ therefore path difference os 0

$p h a s e d i f f e r e n c e = \frac{2 π}{γ} \frac{λ^{2}}{8 d^{2}}$

_{$Y$
A}=a₁cos(wt- $\emptyset$ )

_{$Y$
C}=a₁cos(wt- $\emptyset$ )

Therefore signal picked up by R₂ is

Y_A+Y_b+Y_c+Y_d= y = 2a₁cos(wt- $\emptyset$ )

|y|² = 4a₁²cos²(wt- $\emptyset$ )

=2a₁². Thus R₁ picks up the larger signal.

(ii) if B is swiched off

R₁ picks up y = acoswt

^{IR1=
$\frac{1}{2}$}^a2

R₂ picks up y = a coswt

I_R2= $\frac{1}{2}$ a²

(iii) thus R₁ and R₂ pick up the same signal

If D is switched off

R₁ picks up by y = acoswt

I_R1= $\frac{1}{2}$ a²

Y= 3acoswt

I_R2=9a²/2

R₂ picks up larger signal compared to R₁

(iv) thus a signal at R₁ indicates B has been switched off and an enhanced signal at R₂indicates D has been switched off.

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This is a Long Answer Type Questions as classified in NCERT ExemplarExplanation-consider the disturbance at the receiver R1 which is at a distance d from BYA= acos(wt)  and path difference is <math> <mfrac> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math>  hence phase difference is <math> <mi>π</mi> </math> .Thus the wave R1 because of BYB= acos(wt- <math> <mi>π</mi> </math> )= - acoswt here path difference is <math> <mi>λ</mi> </math>  and hence phase difference is <math> <mn>2</mn> <mi>π</mi> </math> Thus R1 because of CYc= acos(wt-2 <math> <mi>π</mi> </math> )= acoswt(i)let the signal picked up at R2 from B be YB= a1cos(wt)The path difference between signal at D and that B is <math> <mfrac> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> YD= -a1cos(wt)The path difference between signal at A and that atB is <math> <mroot> <mrow> <mrow> <msup> <mrow> <mrow> <mi>d</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mrow> <mo>(</mo> <mfrac> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow></mrow> </mroot> </math> -d = d( <math> <msup> <mrow> <mrow> <mn>1</mn> <mo>+</mo> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> <msup> <mrow> <mrow> <mi>D</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> <mo>/</mo> <mn>2</mn> </mrow> </mrow> </msup> </math> -d = <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>8</mn> <msup> <mrow> <mrow> <mi>d</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> <math> <mi>a</mi> <mi>s</mi> <mi> </mi> <mi>d</mi> <mo>?</mo> <mi>λ</mi> </math>  therefore path difference os 0 <math> <mi>p</mi> <mi>h</mi> <mi>a</mi> <mi>s</mi> <mi>e</mi> <mi> </mi> <mi>d</mi> <mi>i</mi> <mi>f</mi> <mi>f</mi> <mi>e</mi> <mi>r</mi> <mi>e</mi> <mi>n</mi> <mi>c</mi> <mi>e</mi> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>2</mn> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mi>γ</mi> </mrow> </mrow> </mfrac> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>8</mn> <msup> <mrow> <mrow> <mi>d</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> <math> <mi>Y</mi> </math> A=a1cos(wt- <math> <mi>∅</mi> </math> ) <math> <mi>Y</mi> </math> C=a1cos(wt- <math> <mi>∅</mi> </math> )Therefore signal picked up by R2 isYA+Yb+Yc+Yd= y = 2a1cos(wt- <math> <mi>∅</mi> </math> )|y|2 = 4a12cos2(wt- <math> <mi>∅</mi> </math> )=2a12. Thus R1 picks up the larger signal.(ii) if B is swiched offR1 picks up y = acoswtIR1= <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> a2R2 picks up  y = a coswtIR2= <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> a2(iii) thus R1 and R2 pick up the same signalIf D is switched offR1 picks up by y = acoswtIR1= <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> a2Y= 3acoswtIR2=9a2/2R2 picks up larger signal compared to R1(iv) thus a signal at R1 indicates B has been switched off and an enhanced signal at R2 indicates D has been switched off.

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