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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)

best_answer

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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