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SashulF [63]
4 years ago
11

What is the slit spacing of a diffraction necessary for a 600 nm light to have a first order principal maximum at 25.0°?

Physics
1 answer:
Sauron [17]4 years ago
6 0

Explanation:

Given that,

Wavelength of light, \lambda=600\ nm=6\times 10^{-7}\ m

Angle, \theta=25^{\circ}

We need to find the slit spacing for diffraction. For a diffraction, the first order principal maximum is given by :

d\sin\theta=n\lambda

n is 1 here

d is slit spacing

d=\dfrac{\lambda}{\sin\theta}\\\\d=\dfrac{6\times 10^{-7}}{\sin(25)}\\\\d=1.41\times 10^{-6}\ m\\\\d=1.41\ \mu m

So, the slit spacing is 1.41\ \mu m.

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You want to lean your dad's ladder on a smooth wall. If the mass of ladder is 4.42 kg and coefficient
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angle minimum   θ = 41.3º

Explanation:

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    N - W = 0

    N = W

The rotational equilibrium condition, where we place the axis of rotation on the wall

We assume that counterclockwise rotations are positive

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the friction force formula is

     fr = μ N

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we substitute

      μ m g l sin θ - m g l cos θ + mg l /2   cos θ = 0

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       μ sin θ - ½ cos θ = 0

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3 years ago
What can subatomic particles be broken down into?
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An electron with a charge of -1.6 × 10-19 coulombs experiences a field of 1.4 × 105 newtons/coulomb. What is the magnitude of th
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Answer:

Electric force, F=2.24\times 10^{-14}\ N

Explanation:

It is given that,

Charge on an electron is -1.6\times 10^{-19}\ C

Electric field, E=1.4\times 10^5\ N/m

We need to find the magnitude of the electric force on this electron due to this field. The electric force is given by :

F=qE\\\\F=1.6\times 10^{-19}\times 1.4\times 10^5\\\\F=2.24\times 10^{-14}\ N

So, the electric force is 2.24\times 10^{-14}\ N.

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