(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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Consider a two-slit interference arrangement (see figure) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of $D$ in terms of $\lambda$ such that the first minima on the screen falls at a distance $D$ from the centre $O$.

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In $YDSE$,the intensity of the central bright fringe is $8 \, mW/m^2$. What will be the intensity at a path difference of $\frac{\lambda}{6}$?

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In $Y.D.S.E.$ using light of wavelength $\lambda$,the intensity of light at a point on the screen with path difference $\lambda$ is $M$ units. Calculate the intensity of light at a point where the path difference is $\frac{\lambda}{3}$.

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The intensity ratio of the maxima and minima in an interference pattern produced by two coherent sources of light is $9: 1$. The intensities of the light sources used are in the ratio (in $: 1$)

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In Young's double-slit experiment using monochromatic light of wavelength $\lambda$,the intensity of light at a point on the screen where the path difference is $\lambda$ is $K$ units. What is the intensity of light at a point where the path difference is $\lambda / 3$?

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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

Similar Questions

Consider a two-slit interference arrangement (see figure) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of $D$ in terms of $\lambda$ such that the first minima on the screen falls at a distance $D$ from the centre $O$.

In $YDSE$,the intensity of the central bright fringe is $8 \, mW/m^2$. What will be the intensity at a path difference of $\frac{\lambda}{6}$?

In $Y.D.S.E.$ using light of wavelength $\lambda$,the intensity of light at a point on the screen with path difference $\lambda$ is $M$ units. Calculate the intensity of light at a point where the path difference is $\frac{\lambda}{3}$.

The intensity ratio of the maxima and minima in an interference pattern produced by two coherent sources of light is $9: 1$. The intensities of the light sources used are in the ratio (in $: 1$)

In Young's double-slit experiment using monochromatic light of wavelength $\lambda$,the intensity of light at a point on the screen where the path difference is $\lambda$ is $K$ units. What is the intensity of light at a point where the path difference is $\lambda / 3$?

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