1.Light of wavelength 633 nm passes through a narrow slit and forms a diffraction pattern on a screen 8.0 m away. A distance of 23 mm separates the centers of the first minima on either side of the central bright fringe. What is the distance between the fourth minima on both sides of the central bright fringe? My answer is C 2.Light of wavelength 633 nm from a distant source is incident on a slit with a width of 0.750

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chaseashley24

High School

1.Light of wavelength 633 nm passes through a narrow slit and forms a diffraction pattern on a screen 8.0 m away. A distance of 23 mm separates the centers of the first minima on either side of the central bright fringe. What is the distance between the fourth minima on both sides of the central bright fringe?

My answer is C

2.Light of wavelength 633 nm from a distant source is incident on a slit with a width of 0.750 mm wide. A diffraction pattern is observed on a screen 3.50 m away. What is the distance between the dark fringes on either side of the central maximum?
MY ANSWER IS A

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skyluke89

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(a) 184 mm
The formula for the diffraction from single slit is:
$y=\frac{m\lambda D}{a}$
where
y is the distance of the mth-minimum from the central maximum
$\lambda$ is the wavelength
D is the distance of the screen from the slit
a is the width of the slit
In this problem, we have:
$\lambda=633 nm = 633\cdot 10^{-9} m$
D = 8.0 m
$y_1 = 23 mm = 0.023 m$ for m = 1
Solving for a, we find the width of the slit:
$a=\frac{m\lambda D}{y}=\frac{(1)(633\cdot 10^{-9})(8.0)}{0.023}=2.20\cdot 10^{-4} m = 0.22 mm$
Now we can find the distance of the 4th minimum from the central maximum:
$y_4 = \frac{m \lambda D}{a}=\frac{(4)(633\cdot 10^{-9})(8.0)}{2.2\cdot 10^{-4}}=0.092 m = 92 mm$
So, the distance between the forth minima on both sides of the central maximum is
$d=2y_4 = 2(92 mm)=184 mm$
(b) 5.90 mm
Again, we can use the formula
$y=\frac{m\lambda D}{a}$
where in this situation:
$\lambda = 633\cdot 10^{-9} m$
$a=0.750 mm = 7.5\cdot 10^{-4} m$
D = 3.50 m
Solving for m = 1, we find the distance of the first minimum from the central bright fringe:
$y_1=\frac{m\lambda D}{a}=0.00295 m = 2.95 mm$
And so, the distance between the two minima is
$d=2 y_1 = 2(2.95 mm)=5.90 mm$
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