In a Young’s double slit experiment the intensity at a point where the path difference is   \frac{\lambda }{6}    (  \lambda   being the wavelength of light used ) is  I.If I_{0} denotes the maximum intensity, \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}

 

Answers (1)

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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