Two light waves of intensities I and 2I super | vedclass

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(D) Given intensities are $I_1 = I$ and $I_2 = 2I$.Path difference $\Delta x = 12.5 \% 	ext{ of } \lambda = \frac{12.5}{100} \lambda = \frac{\lambda}

Vedclass · Accepted Answer

$5I$

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(D) Given intensities are $I_1 = I$ and $I_2 = 2I$.Path difference $\Delta x = 12.5 \% 	ext{ of } \lambda = \frac{12.5}{100} \lambda = \frac{\lambda}{8}$.Phase difference $\phi = \frac{2\pi}{\lambda} \cdot \Delta x = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{8} = \frac{\pi}{4}$.The resultant intensity $I_R$ is given by the formula $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.Substituting the values: $I_R = I + 2I + 2\sqrt{I \cdot 2I} \cos(\frac{\pi}{4})$.$I_R = 3I + 2\sqrt{2I^2} \cdot \frac{1}{\sqrt{2}}$.$I_R = 3I + 2I \cdot \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 3I + 2I = 5I$.

Two light waves of intensities $I$ and $2I$ superimpose on each other. If the path difference between the light waves reaching a point is $12.5 \%$ of the wavelength of the light,then the resultant intensity at the point is (Both the light waves have same wavelength).

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