(C) The resultant intensity $I$ of two interfering waves with amplitudes $A_1$ and $A_2$ and phase difference $\phi$ is given by $I = A_1^2 + A_2^2 +

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(C) The resultant intensity $I$ of two interfering waves with amplitudes $A_1$ and $A_2$ and phase difference $\phi$ is given by $I = A_1^2 + A_2^2 + 2A_1A_2 \cos \phi$.Given $A_1 = a$,$A_2 = 2a$,and $\phi = \pi$.Substituting these values into the formula:$I = a^2 + (2a)^2 + 2(a)(2a) \cos \pi$.Since $\cos \pi = -1$,we have:$I = a^2 + 4a^2 + 4a^2(-1)$.$I = 5a^2 - 4a^2 = a^2$.Therefore,the resultant intensity is $a^2$.

Class 12 Physics Wave Optics Wave Nature and Interference of Light (Intensity)

Easy

Two light sources have amplitudes $a$ and $2a$ respectively. If they interfere with a phase difference of $\pi$,what is the resultant minimum intensity?

A
$0$
B
$5a^2$
C
$a^2$
D
$9a^2$

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Wave Optics

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Wave Nature and Interference of Light (Intensity)

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Medium

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Two light sources have amplitudes $a$ and $2a$ respectively. If they interfere with a phase difference of $\pi$,what is the resultant minimum intensity?

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Two coherent monochromatic light beams of intensities $I$ and $4I$ are superposed. The maximum and minimum possible intensities in the resulting beam are

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