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Intensity Explanation in YDSE

The ratio of ...

Question

The ratio of the intensities at minima to the maxima in the Young's double slit experiment is 9:25. Find the ratio of the widths of the two slits.

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Solution

If the width of the slits are A₁ and A₂, the amplitude will also be $A_{1}$ and $A_{2}$ .

The maximum amplitude, A' will happen when constructive interference happens-

$A^{'} = (A_{1} + A_{2})$

The maximum amplitude, A will happen when destructive interference happens-

$A = (A_{1} - A_{2})$

We know that intensity is proportional to the square of amplitude.

$\frac{I_{1}}{I_{2}} = \frac{A^{' 2}}{A^{2}} = \frac{(A_{1} + A_{2})^{2}}{(A_{1} - A_{2})^{2}} \frac{9}{25} = (\frac{A_{1} + A_{2}}{A_{1} - A_{2}})^{2} \sqrt{\frac{9}{25}} = \frac{A_{1} + A_{2}}{A_{1} - A_{2}} \frac{A_{1} + A_{2}}{A_{1} - A_{2}} = \frac{3}{5} \frac{A_{1}}{A_{2}} = \frac{5 + 3}{5 - 3} = \frac{8}{2} A_{1} : A_{2} = 4 : 1$

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