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The maximum intensities produced by two ...

The maximum intensities produced by two coherent waves of intensity `I_(1)` and `I_(2)` will be:

A

`I_(1) + I_(2)`

B

`I_(1)^(2) + I_(2)^(2)`

C

`I_(1) + I_(2) + 2sqrt(I_(1)I_(2))`

D

zero

Text Solution

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The correct Answer is:
To find the maximum intensity produced by two coherent waves with intensities \( I_1 \) and \( I_2 \), we can follow these steps: ### Step-by-Step Solution: 1. **Understanding Coherent Waves**: Coherent waves are waves that have a constant phase difference and the same frequency. The intensity of a wave is related to the square of its amplitude. 2. **Relating Intensity to Amplitude**: The intensity \( I \) of a wave is given by the formula: \[ I = A^2 \] where \( A \) is the amplitude of the wave. Therefore, if the intensities of the two waves are \( I_1 \) and \( I_2 \), their amplitudes can be expressed as: \[ A_1 = \sqrt{I_1} \quad \text{and} \quad A_2 = \sqrt{I_2} \] 3. **Net Amplitude Calculation**: When two coherent waves interfere, the resultant amplitude \( A \) can be calculated using the formula: \[ A = A_1 + A_2 + 2 \sqrt{A_1 A_2} \cos(\theta) \] where \( \theta \) is the phase difference between the two waves. 4. **Finding Maximum Intensity**: To find the maximum intensity, we need to maximize the amplitude. The maximum value of \( \cos(\theta) \) is 1 (which occurs during constructive interference). Thus, we set \( \cos(\theta) = 1 \): \[ A_{\text{max}} = A_1 + A_2 + 2 \sqrt{A_1 A_2} \] Substituting the expressions for \( A_1 \) and \( A_2 \): \[ A_{\text{max}} = \sqrt{I_1} + \sqrt{I_2} + 2 \sqrt{\sqrt{I_1} \sqrt{I_2}} = \sqrt{I_1} + \sqrt{I_2} + 2 \sqrt{I_1 I_2} \] 5. **Calculating Maximum Intensity**: The maximum intensity \( I_{\text{max}} \) is given by: \[ I_{\text{max}} = A_{\text{max}}^2 \] Therefore: \[ I_{\text{max}} = \left( \sqrt{I_1} + \sqrt{I_2} + 2 \sqrt{I_1 I_2} \right)^2 \] 6. **Final Expression**: Expanding the square gives: \[ I_{\text{max}} = I_1 + I_2 + 4 \sqrt{I_1 I_2} \] ### Conclusion: The maximum intensity produced by two coherent waves of intensity \( I_1 \) and \( I_2 \) is: \[ I_{\text{max}} = I_1 + I_2 + 4 \sqrt{I_1 I_2} \]
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In YDSE, if sources are incoherent, the intensity on screen is 13 I_(0^(circ)) . When these sources are coherent then minimum intensity on screen is I_(0^(circ)) If maximum intensity produced by these coherent sources on screen is n^(2) I_(0) , then find n .

Find the maxinum intensity in case of interference of n identical waves each of intensity I_(0) if the interference is (a) coherent and (b) incoherent.

Knowledge Check

  • What will be the sum of the minimum and maximum intensities when two periodic coherent waves of intensities I_(1) and I_(2) pass through a region at the same time in the same direction.

    A
    `2 ( l _(1) + l _(2))`
    B
    `I _(1) + I _(2)`
    C
    `( sqrt ( I _(1)) - sqrt ( I _(2)))^(2)`
    D
    `( sqrt ( I _(1)) + sqrt (I _(2)))^(2)`
  • The ratio of maximum intensity to minimum intensity due to interference of two light waves of amplitudes a_(1) and a_(2) is

    A
    `(I_("max"))/(I_("min")) =((a_(1)+a_(2))^(2))/((a_(1)-a_(2))^(2))`
    B
    `(I_("max"))/(I_("min")) =(a_(1)+a_(2))/(a_(1)-a_(2))`
    C
    `(I_("max"))/(I_("min")) =(a_(1)^(2)+a_(2)^(2))/(a_(1)^(2)-a_(2)^(2))`
    D
    none of these
  • Maximum and minimum intensities obtained by two sources having intensities 4I and I are :

    A
    5I , - 3I
    B
    9I , I
    C
    9I, -I
    D
    5I, 3I
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