In a Young’s double slit experiment with light of wavelength\lambda the separation of slits is d and distance of screen is D such that D > > d > >\lambda. If the Fringe width is \beta, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is :

  • Option 1)

    \frac{\beta }{2}

  • Option 2)

    \frac{\beta }{4}

  • Option 3)

    \frac{\beta }{3}

  • Option 4)

    \frac{\beta }{6}

 

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.

 

 

Fringe Width -

\beta = \frac{\lambda D}{d}
 

- wherein

\beta = y_{n+1}-y_{n}

y_{n+1}= Distance of\left ( n+1 \right )^{th}

Maxima= \left ( n+1 \right )\frac{\lambda D}{d}

y_{n}=Distance of n^{th}

 maxima = \frac{n\lambda D}{d}

 

 2I_{0}=4I_{0}cos^{2}\left(\frac{\Delta \phi}{2} \right ) 

Here \Delta \phi = \frac{\pi}{2}

\Delta \phi = \frac{2\pi}{\lambda}\Delta x   so  \Delta x = \frac{\lambda}{4}

\frac{dy}{\Delta}=\frac{\lambda}{4}                                                (i)

\frac{\lambda \Delta}{d}=\beta                                                (ii)

Thefore,    from equation (i) and (ii)

y=\frac{\beta}{4}

Correct option is 2.

 

 


Option 1)

\frac{\beta }{2}

This is an incorrect option.

Option 2)

\frac{\beta }{4}

This is the correct option.

Option 3)

\frac{\beta }{3}

This is an incorrect option.

Option 4)

\frac{\beta }{6}

This is an incorrect option.

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