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solmaris [256]
2 years ago
13

In a single-slit diffraction experiment, the width of the slit through which light passes is reduced. what happens to the width

of the central bright fringe in the resulting diffraction pattern?
Physics
1 answer:
tamaranim1 [39]2 years ago
7 0

The width of the central bright fringe <u>becomes wider</u> in the resulting diffraction pattern of a single-slit diffraction experiment.

<h3>What is diffracted light?</h3>

The act of bending light around corners such that it spreads out and illuminates regions where a shadow is anticipated is known as diffraction of light. In general, since both occur simultaneously, it is challenging to distinguish between diffraction and interference. The diffraction of light is what causes the silver lining we see in the sky. A silver lining appears in the sky when the sunlight penetrates or strikes the cloud.

<h3>What precisely is single slit diffractive?</h3>

The single-slit diffraction experiment allows us to examine the phenomena of light bending, or diffraction, which enables coherent light from a source to interfere with itself and generate the diffraction pattern, a recognizable pattern on the screen. When the sources are small enough to be relative to the wavelength of light, diffraction is seen.

Learn more about diffraction

brainly.com/question/8645206

#SPJ4

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[30 POINTS] An automobile steering wheel is shown. The ideal mechanical advantage of this wheel and axle = _____ .
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Rw/Ra = MA

18cm/2cm= MA

MA = 9

This means that Fi is 1/9 of the force applied to the axil. The distance travelled by Rw is 9 times more than Ri  is that you move 9 times more when turning the wheel using Rw.

Put more simply

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7 0
3 years ago
Coulomb's Law: Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrost
Ahat [919]

Answer:

The magnitude of electrostatic force on each charge is quarter of the magnitude of initial electrostatic force. ( ¹/₄ F)

Explanation:

The electrostatic force between two charges is given by Coulomb's law;

F = \frac{kQ_1Q_2}{r^2}

where;

Q₁ and Q₂ are the magnitude of the charges

r is the distance between the charges

k is Coulomb's constant

Since the charges are identical;

Q₁ = Q

Q₂ = Q

the electrostatic force experienced by each charge is given by;

F =  \frac{kQ^2}{r^2}

When each of the spheres has lost half of its initial charge;

Q₁ = Q/2

Q₂ = Q/2

F_2 = \frac{k(Q/2)(Q/2)}{r^2}\\\\ F_2 = \frac{k(Q)(Q)}{4r^2}\\\\F_2 = \frac{1}{4} (\frac{kQ^2}{r^2} )\\\\F_2 = \frac{1}{4} (F)

Therefore, the magnitude of electrostatic force on each charge is quarter of the magnitude of initial electrostatic force.

6 0
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