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In YDSE, the amplitude of intensity variation of the two sources is found to be 5% of the average intensity. The ratio of the intensities of two interfering sources is
$\left( A \right)2564$
$\left( B \right)1089$
$\left( C \right)1600$
$\left( D \right)800$

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Last updated date: 28th Feb 2024
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Hint:YDSE (Young’s double slit experiment) shows both energy and matter show both wave and particle characteristics. Apply the relation between the maximum intensities and minimum intensities of the two-sources used in Young’s double slit experiment. From the equation we can ratio between the intensities of two interfering sources.

Formula used:
$\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = {\dfrac{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} + 1} \right)}}{{{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} - 1} \right)}^2}}}^2}$
Where ${I_1}$ and ${I_{_2}}$ are the two intensities.

Complete step by step solution:
Two coherent sources of lights placed at a small distance apart are used in Young’s double slit experiment. Usually only magnitudes greater than wavelength of light is used. Young’s double slit experiment helps in understanding the wave theory of light. Commonly used coherent sources in the modern-day experiments is Laser. Young’s double slit experiment firmly establishes that light behaves as a particle and wave.

Now lets us use the formula
$\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = {\dfrac{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} + 1} \right)}}{{{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} - 1} \right)}^2}}}^2}$

Now in the question they have mentioned that the amplitude of intensity variation of two sources is found to be 5% of the average intensity.
Let us assume the average intensity $I$ to be 100. Then ${I_{\max }}$ will be $5\% $ more than the average intensity $I$. Then ${I_{\max }} = 105$ units and ${I_{\min }}$ will be $5\% $ less than the average intensity $I$. ${I_{\min }} = 95$ units.

$\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = {\dfrac{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} + 1} \right)}}{{{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} - 1} \right)}^2}}}^2}$
$\dfrac{{105}}{{95}} = {\dfrac{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} + 1} \right)}}{{{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} - 1} \right)}^2}}}^2}$
We will the value of $\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} $is $40$
Hence$\dfrac{{{I_1}}}{{{I_2}}} = \left( {\dfrac{{{{\left( {40} \right)}^2}}}{{{{\left( 1 \right)}^2}}}} \right)$
$\dfrac{{{I_1}}}{{{I_2}}} = 1600$

Hence option (C) is the right option.

Note: Later they conducted Young’s double slit experiment using electrons and the pattern generated a similar result as light. It behaves both as a particle and wave. Young’s double slit experiment shows both these characteristics prominently.