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(D) The ratio of maximum intensity to minimum intensity is given by $\frac{I_{\max}}{I_{\min}} = 9$.We know that $\frac{I_{\max}}{I_{\min}} = \left( \

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Both $(b)$ and $(c)$

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(D) The ratio of maximum intensity to minimum intensity is given by $\frac{I_{\max}}{I_{\min}} = 9$.We know that $\frac{I_{\max}}{I_{\min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2 = 9$.Taking the square root on both sides,we get $\frac{a_1 + a_2}{a_1 - a_2} = 3$.Solving for the ratio of amplitudes: $a_1 + a_2 = 3a_1 - 3a_2 \Rightarrow 4a_2 = 2a_1 \Rightarrow \frac{a_1}{a_2} = 2$.Since intensity $I \propto a^2$,the ratio of intensities of individual sources is $\frac{I_1}{I_2} = \left( \frac{a_1}{a_2} \right)^2 = 2^2 = 4$,which means $I_1 : I_2 = 4 : 1$.Thus,both statements $(b)$ and $(c)$ are correct.

In the Young's double slit experiment,the ratio of intensities of bright and dark fringes is $9$. This means that

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