# In a Young’s double slit experiment the intensity at a point where the path difference is   $\dpi{100} \frac{\lambda }{6}$    (  $\dpi{100} \lambda$   being the wavelength of light used ) is  $\dpi{100} I$.If $\dpi{100} I_{0}$ denotes the maximum intensity, $\dpi{100} \frac{I}{I_{0}}$ is equal to Option 1) $\frac{3}{4}$ Option 2) $\frac{1}{\sqrt{2}}$ Option 3) $\frac{\sqrt{3}}{2}$ Option 4) $\frac{1}{2}$

As we learnt in

Malus Law -

$I= I_{0}\cdot \cos ^{2}\theta$

$\theta =$ angle made by E vector with transmission axis.

- wherein

$I=$ Intensity of transmitted light after polarisation .

$I_{0}=$ Intensity of  incident light.

In Young’s double slit experiment intensity at a point is given by

$I=I_{0}\cos ^{2}\left ( \frac{\phi }{2} \right )$

where $\phi =$ phase difference,  $I_{0}=$ maximum intensity

or        $\frac{I}{I_{0}}= \cos ^{2}\left ( \frac{\phi }{2} \right )\cdots \cdots \cdots \cdots (i)$

Phase difference $\phi = \frac{2\pi }{\lambda }\times path \; di\! f\! \! f\! erence$

$\therefore \; \; \; \; \; \; \; \phi = \frac{2\pi }{\lambda }\times \frac{\lambda }{6}\; \; \; \; or\; \; \; \phi = \frac{\pi }{3}\cdots \cdots \cdots \cdots (ii)$

Substitute eq (ii) in eqn (i) we get

$\frac{I}{I_{0}}= \cos ^{2}\left ( \frac{\pi }{6} \right )\; \; \; or\: \: \: \frac{I}{I_{0}}= \frac{3}{4}$

Correct option is 1.

Option 1)

$\frac{3}{4}$

This is the correct option.

Option 2)

$\frac{1}{\sqrt{2}}$

This is an incorrect option.

Option 3)

$\frac{\sqrt{3}}{2}$

This is an incorrect option.

Option 4)

$\frac{1}{2}$

This is an incorrect option.

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