In YDSE,let S_1 and S_2 be the two slits,and | vedclass

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(A) Let $d$ be the distance between the slits $S_1$ and $S_2$,and $D$ be the distance between the slits and the screen.From the geometry of the setup,

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$\frac{\lambda}{	heta}$

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(A) Let $d$ be the distance between the slits $S_1$ and $S_2$,and $D$ be the distance between the slits and the screen.From the geometry of the setup,the angle $	heta$ subtended by the slits at the centre of the screen $C$ is given by $	an(	heta/2) = \frac{d/2}{D}$.For small angles,$	an(	heta/2) \approx 	heta/2 = \frac{d}{2D}$,which implies $	heta = \frac{d}{D}$.The fringe width $\beta$ is defined as $\beta = \frac{\lambda D}{d}$.Substituting $d/D = 	heta$ into the expression for fringe width,we get $\beta = \frac{\lambda}{	heta}$.

In $YDSE$,let $S_1$ and $S_2$ be the two slits,and $C$ be the centre of the screen. If $\theta$ is the angle $S_1CS_2$ and $\lambda$ is the wavelength,the fringe width will be

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