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In a Young's double slit experiment at a point P on the screen the light from two slit with a path difference of $\large\frac{\lambda}{6} $ Where $\lambda$ is the wavelength of the incident light . The intensity of light at point P is I. If $I_0$ is maximum intensity then $\large\frac{I}{I_0}$

$(a)\;\frac{1}{\sqrt 2} \\ (b)\;\frac{\sqrt 3}{2} \\ (c)\;\frac{1}{2} \\ (d)\;\frac{3}{4} $

1 Answer

$I= I_0 \cos^2 (\large\frac{\phi}{2})$
Where $\phi$ is phase difference .
Also phase difference $\phi =\large\frac{2 \pi}{\lambda} $$\times path\; differences$
$\qquad= \large\frac{2 \pi}{\lambda} \times \large\frac{\lambda}{6}=\frac{\pi}{3}$
$\therefore \large\frac{I}{I_0} =\cos^2 \bigg(\large\frac{ \phi}{2}\bigg)$
$\qquad= \cos^2 \bigg( \large\frac{\pi}{6} \bigg)$
$\qquad= \bigg( \large\frac{\sqrt 3}{2} \bigg)^2$
$\qquad= \large\frac{3}{4}$
Hence d is the correct answer.
answered Jan 23, 2014 by meena.p

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