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Zolol [24]
3 years ago
10

In a Young's double-slit experiment the separation distance y between the second-order bright fringe and the central bright frin

ge on a flat screen is 0.0176 m, when the light has a wavelength of 425 nm. Assume that the angles are small enough so that is approximately equal to . Find the separation y when the light has a wavelength of 614 nm.
Physics
1 answer:
atroni [7]3 years ago
8 0

Answer:

Y = 0.0254 m = 25.4 mm

Explanation:

The formula for the fringe spacing in Young's Double-slit experiment is given by the following formula:

Y = \frac{\lambda L}{d}

where,

Y = fringe spacing = 0.0176 m

λ = wavelength = 425 nm = 4.25 x 10⁻⁷ m

L = Distance between screen and slits

d = slit separation

Therefore,

\frac{0.0176\ m}{4.25\ x\ 10^{-7}\ m} = \frac{L}{d}\\\\\frac{L}{d} =  41411.76

Now, for:

λ = 614 nm = 6.14 x 10⁻⁷ m

Y = (6.14\ x\ 10^{-7}\ m)(41411.76)\\

<u>Y = 0.0254 m = 25.4 mm</u>

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Answer: 5.72 x 10-3Ω

Explanation:

Hi, to answer this question, first we have to calculate the cross sectional area of the cable:

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Since: 1000mm = 1m

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Resistance formula:

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Replacing with the values given:

R = (2.82x10-8 x 5.87) / 0.000028937

R = 5.72 x 10-3Ω

Feel free to ask for more if needed or if you did not understand something.

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Explanation:

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