Width of one of the two slits in a Young's double slit interference...

Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is :

Options

A $(2 \sqrt{2}+1):(2 \sqrt{2}-1)$
B $(3+2 \sqrt{2}):(3-2 \sqrt{2})$
C $9: 1$
D $3: 1$

Correct answer

B. $(3+2 \sqrt{2}):(3-2 \sqrt{2})$

Step-by-step solution

\mathrm{I} \propto

width

\quad\quad

\mathrm{I}

_{\max }=\left(\sqrt{

\mathrm{I}

_1}+\sqrt{

\mathrm{I}

_2}\right)^2

\therefore

\mathrm{I}

_1=

\mathrm{I}

_0,

\mathrm{I}

_2=2

\mathrm{I}

_0 \quad\quad

\mathrm{I}

_{\min }=\left(\sqrt{

\mathrm{I}

_1}-\sqrt{

\mathrm{I}

_2}\right)^2

\frac{I_{\max }}{I_{\min }}=\frac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)^2} \Rightarrow \frac{3+2 \sqrt{2}}{3-2 \sqrt{2}}

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