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Andrej [43]
2 years ago
12

A single slit of width 0.50 mm is illuminated with light of wavelength 500 nm, and a screen is placed 120 cm in front of the sli

t. Find the widths (lengths, not angles)of(a) the central maximum,(b) the first maximum beyond the central maximum, and(c) the second maximum beyond the central maximum.
Physics
1 answer:
Vaselesa [24]2 years ago
4 0

Answer:

a) 2.4 mm

b) 1.2 mm

c) 1.2 mm

Explanation:

To find the widths of the maxima you use the diffraction condition for destructive interference, given by the following formula:

m\lambda=asin\theta

a: width of the slit

λ: wavelength

m: order of the minimum

for little angles you have:

y=\frac{m\lambda D}{a}

y: height of the mth minimum

a) the width of the central maximum is 2*y for m=1:

w=2y_1=2\frac{1(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}=2.4*10^{-3}m=2.4mm

b) the width of first maximum is y2-y1:

w=y_2-y_1=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[2-1]=1.2mm

c) and for the second maximum:

w=y_3-y_2=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[3-2]=1.2mm

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What will be the ME of a machine that produces a 240 N work with a 300
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Given the following data;

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where F is force, m is mass and a is acceleration

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Use the mass spectrum of europium to determine the atomic mass of europium. where the peak representing eu-151 has an exact mass
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From the given information:

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Abundance of Eu-151 atom:

X_{Eu-151}=\frac{0.9161}{0.9161+1.000}=0.4781

Abundance of Eu-153 atom:

X_{Eu-153}=\frac{1.000}{0.9161+1.000}=0.5219

Atomic mass of Europium atom:

A=(X_{Eu-151}\times150.91986+X_{Eu-153}\times152.92138)amu\\A=(0.4781\times150.91986+0.5219\times152.92138)amu=151.96445 amu

Therefore, the atomic mass of a Europium atom is 151.96445 amu.

3 0
3 years ago
Read 2 more answers
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