In a double-slit experiment, the third-order maximum for light of wavelength 510 nm is located 17 mm from the central bright spot on a screen 1.6 m from the slits. Light of wavelength 670 nm is then projected through the same slits. Part A How far from the central bright spot will the second-order maximum of this light be located

Answer:

14.9 mm

Explanation:

We know dsinθ = mλ where d = separating of slit, m = order of maximum = 3 and λ = wavelength = 510nm = 510 × 10⁻⁹ m

Also tanθ = L/D where L = distance of m order fringe from central bright spot = 17 mm = 0.017 m and D = distance of screen from slit = 1.6 m

So, sinθ = mλ/d

Since θ is small, sinθ ≅ tanθ

So,

mλ/d = L/D

d = mλD/L

Substituting the values of the variables into the equation, we have

d = 3 × 510 × 10⁻⁹ m × 1.6 m/0.017 m

d = 2448 × 10⁻⁹ m²/0.017 m

d = 144000 × 10⁻⁹ m

d = 1.44 × 10⁻⁴ m

d = 0.144 × 10⁻³ m

d = 0.144 mm

Now, for the second-order maximum, m' of the 670 nm wavelength of light,

m'λ'/d = L'/D where m' = order of maximum = 2, λ' = wavelength of light = 670 nm = 670 × 10⁻⁹ m, d = slit separation = 0.144 mm = 0.144 × 10⁻³ m, L' = distance of second order maximum from central bright spot and D = distance of screen from slit = 1.6m

So, L' = m'λ'D/d

So, substituting the values of the variables into the equation, we have

L' = 2 × 670 × 10⁻⁹ m × 1.6 m/0.144 × 10⁻³ m

L' = 2144  × 10⁻⁹ m²/0.144 × 10⁻³ m

L' = 14888.89 × 10⁻⁶ m

L' = 0.01488 m

L' ≅ 0.0149 m

L' = 14.9 mm

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What did scientists discover?
A. The universe has 73% hydrogen and 25% helium, which does not
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B. Planets are made mostly of other elements, which does not
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C. Stars are made mostly of other elements, which supports the big
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D. The universe has 74% hydrogen and 24% helium, which supports
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a. 0.0014 meters
b. 0.45 meters
c. 2.2 meters
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