In Young's double slit experiment, the intensities of dark and bright fringes are $I$ and $4 I$ respectively. The ratio of amplitudes of sources is

4 : 1

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1 : 3

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3 : 1

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1 : 2

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Solution

The correct option is C $3 : 1$
$A = \sqrt{I}$
$\frac{4}{1} = \frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}}$
=> $\frac{4}{1} = \frac{A_{1} + A_{2}}{A_{1} - A_{2}}$
=> $\frac{A_{1}}{A_{2}} = \frac{3}{1}$

Answer. C) 3:1

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In Young's double slit experiment, the intensities of dark and bright fringes are I and 4I respectively. The ratio of amplitudes of sources is

The correct option is C 3:1A=√I41=√I1+√I2√I1−√I2=> 41=A1+A2A1−A2=>A1A2=31Answer. C) 3:1

In Young's double slit experiment, the intensities of dark and bright fringes are $I$ and $4 I$ respectively. The ratio of amplitudes of sources is

The correct option is C $3 : 1$
$A = \sqrt{I}$
$\frac{4}{1} = \frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}}$
=> $\frac{4}{1} = \frac{A_{1} + A_{2}}{A_{1} - A_{2}}$
=> $\frac{A_{1}}{A_{2}} = \frac{3}{1}$

Answer. C) 3:1