Question

please answer question d and e only, thank you

The picture shows the interference pattern produced by a multiple slit experiment. The distance between the slits and the screen is L, the wavelength is 2. The slits are separated by a distance d and each have a width of a. You may use the small angle approximation. 2=650nm d = 0.125 mm Sam 4 6 5 7 9 8 2 3 10 11 12 14 13 15 www.PrintableRulers net a. How many slits were used in this experiment? No explanation required. A. 1 B. 2 C.3 D.4 E.5 b. Find L, the distance between the slits and the screen. Indicate in the picture what data you are using. Give your answer to three significant figures. c. Find the slit width a. a Sam 5 6 7 8 9 10 11 12 14 13 15 www Printable Rulers.net Id. In the figure above, the point labeled 'a' on the screen is the nimum at 8.5 cm. For this point, what is the path length difference AD ad; for two adjacent slits? Express your answer in units of nanometers. e. In the figure above, mark • Two principal maxima, and label them 'b'. Two minima that are caused by multiple slit interference and NOT by single slit diffraction, label them'c. Three minima that are due to single slit diffraction, and label them'd

Answer 1

d. There are 5 slits, the the phase difference between two consecutive slits be $\delta$ and amplitude of light from individual slits be a.

Then amplitude of light due to all slits

$A=a(1+e^{i\delta}+e^{i2\delta}+e^{i3\delta}+e^{i4\delta})$

Central maxima is at 7cm, for that $\delta=0$

For primary maxima, just before 'a', (As it is a primary maxima, light from all slits should be in phase)

27

For point 'a',$\delta$ should be greater than

211

amplitude=0

$a(1+e^{i\delta}+e^{i2\delta}+e^{i3\delta}+e^{i4\delta})=0$

$a\frac{1-e^{i5\delta}}{1-e^{i\delta}}=0$

$e^{i5\delta}=1$

${5\delta}=2n\pi$

${\delta}=\frac{2n\pi}{5}$

As, $\delta$ should be just greater than

211

So, the very first value after
211
is ${\delta}=\frac{12\pi}{5}$

So, the path difference, $\Delta D_{adj}=\frac{12\pi}{5}*\frac{\lambda}{2\pi}=\frac{6\lambda}{5}$

$\Delta D_{adj} =\frac{6*650nm}{5}=780nm$

e.

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