Question 108 of 191

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 the maximum intensity on either side is:

A$\dfrac{\beta }{4}$B$\dfrac{\beta }{3}$C$\dfrac{\beta }{6}$D$\dfrac{\beta }{2}$

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Answer: (A) $\dfrac{\beta }{4}$

As ${I}_{max}=4I$

Then ${I}_{net}=\dfrac{1}{2}{I}_{max}=2I$

$\Rightarrow 2I=2I\left(1+cos\phi \right)$

$\Rightarrow 1+cos\phi =1$

$\Rightarrow cos\phi =0$

$\Rightarrow \phi =\dfrac{\pi }{2},\dfrac{3\pi }{2},\dfrac{5\pi }{2},\ldots$

$\phi =\dfrac{2\pi }{\lambda }\Delta x$

$\Rightarrow \dfrac{\pi }{2}=\dfrac{2\pi }{\lambda }\Delta x$

$\Rightarrow \Delta x=\dfrac{\lambda }{4},\dfrac{3\lambda }{4},\dfrac{5\lambda }{4},\ldots$



Also path difference $\Delta x=d\sin ⁡\theta =\dfrac{\lambda }{4}$

At desired location

$\Rightarrow \dfrac{d{y}_{n}}{D}=\dfrac{\lambda }{4}$

$\Rightarrow {y}_{n}=\dfrac{\lambda D}{4d}=\dfrac{\beta }{4}\left\{∵\beta =\dfrac{\lambda D}{d}\right\}$

$\Rightarrow$ Separation between central maxima and desired $y=\dfrac{\beta }{4}$