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Lunna [17]
3 years ago
9

A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the

width of the central bright spot on the screen change if the wavelength of the illuminating light is doubled
Physics
1 answer:
sweet [91]3 years ago
5 0

Answer:

Doubling the wavelength of the diffracting doubles the angle of diffraction. So, the width of the central bright spot pattern formed on the screen will also be doubled.

Explanation:

For a single slit diffraction, the path length difference is related to the wavelength of the light leaving the slit onto the screen by

D sin θ = mλ

where D sin θ is the path length of the waves, each.

mλ is the wavelength of the wavelet

where m is the the order of each minimum

m = m = 1,−1,2,−2,3, . . .

The wavelength of each wavelet is always a multiple of the wavelength of the light source, and from the equation, we can see that the angle of diffraction depend on the wavelength of the light. From this we can see that increasing the wavelength of the light increases the angle of diffraction, and hence we can say that doubling the wavelength will double the diffraction angle. Also, the width of the central bright spot of the screen will spread or increase with the angle of diffraction, so doubling the wavelength doubles the central bright spot on the screen.

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You are conducting an experiment inside an elevator that can move in a vertical shaft. A load is hung vertically from the ceilin
Ivan

Answer: The elevator must be accelerating.

Explanation:

As the tension force is opposing to the the force of gravity on the load which is hung vertically, and the tension force can adopt any value in order to comply with Newton's 2nd law, if the tension force is less than the force due to gravity, this means that all system is not in equilibrium, so it must be accelerating.

If we assume that the downward is the positive direction, we can write:

mg - T = ma

If T = 0.9 mg, ⇒ mg (1-0.9) =0.1 mg = m a ⇒a = 0.1 g , in downward direction.

5 0
3 years ago
Inside a vacuum tube, an electron is in the presence of a uniform electric field with a magnitude of 320 N/C. (a) What is the ma
nignag [31]

(a) The magnitude of the acceleration of the electron is 5.62 x 10¹³ m/s².

(b) The speed of the electron after the given time is  4.78 x 10⁵ m/s.

<h3>Acceleration of the electron</h3>

The acceleration of the electron is calculated as follows;

F = qE

ma = qE

a = qE/m

a = (1.6 x 10⁻¹⁹ x 320)/(9.11 x 10⁻³¹)

a = 5.62 x 10¹³ m/s²

<h3>Speed of the electron</h3>

v = at

v = 5.62 x 10¹³ m/s² x  8.50 x 10⁻⁹ s

v = 4.78 x 10⁵ m/s

Learn more about speed here: brainly.com/question/4931057

#SPJ1

7 0
2 years ago
A 0.150 kg ball on the end of a 1.10 m long cord (negligible mass)is swung in a vertical circle..
Aneli [31]
<span> For any body to move in a circle it requires the centripetal force (mv^2)/r. In this case a ball is moving in a vertical circle swung by a mass less cord. At the top of its arc if we draw its free body diagram and equate the forces in radial direction to the centripetal force we get it as T +mg =(mv^2)/r T is tension in cord m is mass of ball r is length of cord (radius of the vertical circle) To get the minimum value of velocity the LHS should be minimum. This is possible when T = 0. So minimum speed of ball v at top =sqrtr(rg)=sqrt(1.1*9.81) = 3.285 m/s In the second case the speed of ball at top = (2*3.285) =6.57 m/s Let us take the lowest point of the vertical circle as reference for potential energy and apllying the conservation of energy equation between top & bottom we get velocity at bottom as 9.3m/s. Now by drawing the free body diagram of the ball at the bottom and equating the net radial force to the centripetal force T-mg=(mv^2)/r We get tension in cord T=13.27 N</span>
3 0
3 years ago
Read 2 more answers
Most of the Earth's volcanoes occur
timofeeve [1]

Answer:

A.

Explanation:

NEAR THE CENTER OF TECTONIC PLATES.

3 0
3 years ago
A 0.43 kg hammer is moving horizontally at 5.2 m/s when it strikes a nail and comes to rest after driving the nail 0.014 m into
vagabundo [1.1K]

Answer:

The duration of the impact is 0.005384 seconds

Explanation:

Given

m = 0.43 kg

v = 5.2 m/s

x = 0.014 m

Knowing the formulas

v^{2}_{f} = v^{2}_{i} + 2ax\\0 = 5.2^{2}  + 2a*0.014\\a = - 965.71 m/s^{2} \\\\vf = vi + at\\0 = 5.2 + (965.71)t\\t = 0.005384 s

5 0
3 years ago
Read 2 more answers
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