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In Young's double slit experiment, the interference pattern is found to have an intensity ratio between bright and dark fringes is $9 .$ This implies the
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the intensities at the screen due to the two slits are 4 units and 1 units respectively
For bright fringes, $\left.\right|_{\max }=(a+b)^{2}$
For dark fringes, $I_{\min }=(a-b)^{2}$
Now $\frac{\mathrm{I}_{\max }}{\mathrm{I}_{\min }}=9=\frac{(\mathrm{a}+\mathrm{b})^{2}}{(\mathrm{a}-\mathrm{b})^{2}} \Rightarrow \frac{\mathrm{a}+\mathrm{b}}{\mathrm{a}-\mathrm{b}}=3$
$\Rightarrow a+b=3(a-b) \Rightarrow a+b=3 a-3 b \Rightarrow 2 a=4 b$
$\frac{\mathrm{a}}{\mathrm{b}}=2 \Rightarrow \frac{\mathrm{a}^{2}}{\mathrm{~b}^{2}}=4$
$\therefore$ Ratio of intensities of the two slits,
$\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{\mathrm{a}^{2}}{\mathrm{~b}^{2}}=\frac{4}{1}$
For dark fringes, $I_{\min }=(a-b)^{2}$
Now $\frac{\mathrm{I}_{\max }}{\mathrm{I}_{\min }}=9=\frac{(\mathrm{a}+\mathrm{b})^{2}}{(\mathrm{a}-\mathrm{b})^{2}} \Rightarrow \frac{\mathrm{a}+\mathrm{b}}{\mathrm{a}-\mathrm{b}}=3$
$\Rightarrow a+b=3(a-b) \Rightarrow a+b=3 a-3 b \Rightarrow 2 a=4 b$
$\frac{\mathrm{a}}{\mathrm{b}}=2 \Rightarrow \frac{\mathrm{a}^{2}}{\mathrm{~b}^{2}}=4$
$\therefore$ Ratio of intensities of the two slits,
$\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{\mathrm{a}^{2}}{\mathrm{~b}^{2}}=\frac{4}{1}$
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