Two monochromatic light beams of intensity 16 | vedclass

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(D) Intensity $I \propto a^2$,where $a$ is the amplitude.Given $I_1 = 16$ and $I_2 = 9$.Thus,the ratio of amplitudes is $\frac{a_1}{a_2} = \sqrt{\frac

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$49/1$

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(D) Intensity $I \propto a^2$,where $a$ is the amplitude.Given $I_1 = 16$ and $I_2 = 9$.Thus,the ratio of amplitudes is $\frac{a_1}{a_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{16}{9}} = \frac{4}{3}$.Let $a_1 = 4k$ and $a_2 = 3k$.The intensity of bright fringes is $I_{max} = (a_1 + a_2)^2 = (4k + 3k)^2 = (7k)^2 = 49k^2$.The intensity of dark fringes is $I_{min} = (a_1 - a_2)^2 = (4k - 3k)^2 = (k)^2 = k^2$.The ratio of intensities is $\frac{I_{max}}{I_{min}} = \frac{49k^2}{k^2} = \frac{49}{1}$.

Two monochromatic light beams of intensity $16$ and $9$ units are interfering. The ratio of intensities of bright and dark parts of the resultant pattern is

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