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A double slit aperture is illuminated by light of

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a) The distance from the central bright fringe and the second central bright fringe is

y=R\frac{\lambda}{d}.......(that equation is only valid for smalls angles)

y=(5.00m)\frac{530\times 10^{-9}m}{0.1\times 10^{-3}m}

y=0.0265m

y=2.65cm

b) The destructive interference is given by the equation

\sin(\theta)=\left((0)+\frac{1}{2} \right )\frac{\lambda}{d}

\sin(\theta)=\left(\frac{1}{2} \right )\frac{\lambda}{d}

\sin(\theta)=\frac{\lambda}{2d}

and we can find the angle

\theta=\arcsin\left(\frac{\lambda}{2d}\right)

\theta=\arcsin\left(\frac{(530\times 10^{-9}m)}{2(0.1\times 10^{-3}m)}\right)

\theta=0.152^o

The distance from the center to the first dark fringer is

y'=R\tan(\theta)

y'=(5.00m)\tan(0.152^o)

y'=0.0133m

Then the distance between the two dark fringes is

y_2=2y'

y_2=2(0.0133m)

y_2=0.0266m

y_2=2.66cm

c) The diffraction pattern is given by only one slit. In the diffraction pattern the first dark fringe is given by

\sin(\theta)=\frac{\lambda}{a}

\theta=\arcsin\left( \frac{\lambda}{a}\right )

\theta=\arcsin\left( \frac{530\times 10^{-9}m}{2.125\times 10^{-6}m}\right )

\theta=14.4^o

As before the the distance between the dark fringes is twice the distance from the center one of the fringes

y=2y'

y=2R\tan(\theta)

R=\frac{y}{2\tan(\theta)}

R=\frac{1m}{2\tan(14.4^o)}

R=1.95m

d) In the laboratory you can see that the dark fringe in not a line. The drak fringe has width that can be not neglegible. The previous equation give the distance from the center of one fringe to the center of the other, not the widht of the central peak of the diffraction pattern

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