(B) Given: Intensities of two beams of monochromatic light are $I_1 = 64 \ mW$ and $I_2 = 4 \ mW$.The formula for the resultant intensity of two inter

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(B) Given: Intensities of two beams of monochromatic light are $I_1 = 64 \ mW$ and $I_2 = 4 \ mW$.The formula for the resultant intensity of two interfering waves is given by $I = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos \phi$.When the intensity is reduced to $84 \ mW$,we substitute the known values into the equation:$84 = 64 + 4 + 2 \sqrt{64 \times 4} \cos \phi$$84 = 68 + 2 \times 8 \times 2 \cos \phi$$84 - 68 = 32 \cos \phi$$16 = 32 \cos \phi$$\cos \phi = \frac{16}{32} = \frac{1}{2}$Therefore,$\phi = \cos^{-1} \left( \frac{1}{2} \right) = 60^\circ$.

Class 12 Physics Wave Optics Intensity in Young's Double Slit Experiment

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Two beams of monochromatic light with intensities $64 \ mW$ and $4 \ mW$ interfere constructively to produce an intensity of $100 \ mW$. If one of the beams is shifted by a phase angle $\phi$,the intensity is reduced to $84 \ mW$. The magnitude of $\phi$ is

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A
$30^\circ$
B
$60^\circ$
C
$45^\circ$
D
$\cos^{-1} \left( \frac{1}{3} \right)$

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Wave Optics

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Intensity in Young's Double Slit Experiment

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If the intensity of each wave in the observed interference pattern in Young's double slit experiment is $I_0$,then for some point $P$ where the phase difference is $\phi$,the resultant intensity $I$ will be:

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In Young's double slit experiment,the intensity of light coming from the first slit is double the intensity from the second slit. The ratio of the maximum intensity to the minimum intensity on the interference fringe pattern observed is

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Two beams of monochromatic light with intensities $64 \ mW$ and $4 \ mW$ interfere constructively to produce an intensity of $100 \ mW$. If one of the beams is shifted by a phase angle $\phi$,the intensity is reduced to $84 \ mW$. The magnitude of $\phi$ is

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