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In a Young's double slit experiment the intensity of light when slit is at distance $\lambda$ from central is I. What will be the intensity at the distance of slit is $\frac{\lambda}{6}$ ?
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The correct answer is:
$\frac{3}{4} I$
As, $\phi=\frac{2 \pi}{\lambda} \Delta x \Rightarrow \phi=\frac{2 \pi}{\lambda} \lambda$
$\phi=2 \pi$
$\therefore I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi$
$=I^{\prime}+I^{\prime}+2 I^{\prime} \cos 2 \pi$
$I=2 I^{\prime}+2 I^{\prime}=4 I^{\prime}$
When $\Delta x=\frac{\lambda}{6}$, then
$\phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{6}=\frac{2 \pi}{6}$
$\therefore \quad I_1=I^{\prime}+I^{\prime}+2 \sqrt{I^{\prime} I^{\prime}} \cos \frac{2 \pi}{6}$
$I_1=2 I^{\prime}+2 I^{\prime}(1 / 2)$
$I_1=3 I^{\prime}$
$\therefore \quad \frac{I_1}{I}=\frac{3 I^{\prime}}{4 I^{\prime}}=\frac{3}{4}$
or $\quad I_1=\frac{3}{4} I$
$\phi=2 \pi$
$\therefore I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi$
$=I^{\prime}+I^{\prime}+2 I^{\prime} \cos 2 \pi$
$I=2 I^{\prime}+2 I^{\prime}=4 I^{\prime}$
When $\Delta x=\frac{\lambda}{6}$, then
$\phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{6}=\frac{2 \pi}{6}$
$\therefore \quad I_1=I^{\prime}+I^{\prime}+2 \sqrt{I^{\prime} I^{\prime}} \cos \frac{2 \pi}{6}$
$I_1=2 I^{\prime}+2 I^{\prime}(1 / 2)$
$I_1=3 I^{\prime}$
$\therefore \quad \frac{I_1}{I}=\frac{3 I^{\prime}}{4 I^{\prime}}=\frac{3}{4}$
or $\quad I_1=\frac{3}{4} I$
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