Question

A double-slit interference pattern is created by two narrow slits spaced 0.30 mm apart. The distance between the first and the fifth minimum on a screen 40 cm behind the slits is 6 mm. What is the wavelength of the light used in this experiment?

Answer 1

The formulas to solve this problem are:

$dsin\Theta _{dark}=(m+\frac{1}{2})\lambda$

$y=Ltan\Theta$

but there is a exception to this formulas, when the angle is too small we can assume:

$y=L(\frac{m\lambda }{d})$

This is true since the linear nature of the pattern.

Now for minimums:

$y=L(\frac{(m+\frac{1}{2})\lambda }{d})$

We have a delta y:

$\Delta y=6mm$

$6mm=L(\frac{(m_{5}+\frac{1}{2})\lambda }{d})-L(\frac{(m_{1}+\frac{1}{2})\lambda }{d})$

We substitute the "m"

$6mm=L(\frac{(5+\frac{1}{2})\lambda }{d})-L(\frac{(1+\frac{1}{2})\lambda }{d})$

$6mm=L(\frac{5.5\lambda }{d})-L(\frac{1.5\lambda }{d})$

We proceed to clear Lambda:

$6mm=\frac{L}{d}(5.5\lambda-1.5\lambda)$

$6mm\frac{d}{L}=4\lambda$

Our final equation:

$\lambda =6*10^{-3}m\frac{d}{4L}=6*10^{-3}m\frac{(0.3*10^{-3}m)}{4(40*10^{-2}m)}=1.125*10^{-6}m=1.125\mu m$

Good luck!