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Setler79 [48]
2 years ago
12

Monochromatic light with a wavelength of 600 nanometers (one nanometer is 10-9 meters) is incident upon a double slit arrangemen

t. The interference fringes are seen on a screen that is 10 m from the slits. If the bright fringes are 5 mm apart, what is the slit separation; that is, what is the value of d? Hint: convert all of the values into meters first.

Physics
1 answer:
AnnyKZ [126]2 years ago
7 0
I think it should be 6,000
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a)

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Therefore,

(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0.8\ m/s)+(1.8\ kg)(v_2)\\v_2 = 5\ m/s - 0.8\ m/s

<u>v₂ = 4.2 m/s</u>

<u></u>

b)

We will use the law of conservation of momentum here:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

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u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0 m/s

v₂ = speed of second after before collsion = ?

Therefore,

(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0\ m/s)+(1.8\ kg)(v_2)

<u>v₂ = 5 m/s</u>

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