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Varvara68 [4.7K]

3 years ago

9

two charges separated a distance of 1.0 meter exert a 1.0-N force on each other If the charges are pushed to a separation of 1/4

meter the force of each charge would be?

1 answer:

Effectus [21]3 years ago

8 0

Newtons of force is 20

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Answer: 117.8 nm

Explanation:

Given,

Nonreflective coating refractive index : n = 1.21

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Wave length of light = λ = 570 nm = $570\times10^{-9}\ m$

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$=\dfrac{570\times10^{-9}\ m}{4\times1.21}\\\\\approx\dfrac{117.8\times 10^{-9}\ m}{1}\\\\=117.8\text{ nm}$

Hence, the minimum thickness of the coating that will accomplish= 117.8 nm

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