In the Young's double slit experiment, the interference pattern is found to have an intensity ratio between the bright and dark fringes as 25/9. This implies that

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the intensities at the screen due to each of the two slits are 16 units and 1 units respectively

the intensities at the screen due to the two slits are 25 units and 9 units respectively

the amplitude ratio is 5/3

the amplitude ratio is 4

answer is D, A.

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Detailed Solution

$\frac{I_{\max}}{I_{\min}} = \frac{{(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}}{{(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}} = \frac{9}{1} \begin{array}{l} \Rightarrow \frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}} = \frac{3}{1} \Rightarrow 4 \sqrt{I_{2}} = 2 \sqrt{I_{1}} \\ \Rightarrow \sqrt{\frac{I_{1}}{I_{2}}} = \frac{4}{2} \Rightarrow \frac{I_{1}}{I_{2}} = 4 = \frac{a^{2}}{b^{2}} \\ \Rightarrow \frac{a}{b} = 2 \end{array}$