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Nostrana [21]
3 years ago
8

A double-slit diffraction pattern is formed on a distant screen. If the separation between the slits decreases, what happens to

the distance between interference fringes? Assume the angles involved remain small.The distance between interference fringes remains the same.The effect cannot be determined unless the distance between the slits and the screen is known.The distance between interference fringes also decreases.The distance between interference fringes increases.
Physics
1 answer:
LenKa [72]3 years ago
7 0

Answer:

The distance between interference fringes increases.

Explanation:

In a double-slit diffraction pattern, the angular position of the nth-maximum in the diffraction patter (measured with respect to the central maximum) is given by

sin \theta = \frac{n \lambda}{d}

where

\theta is the angular position

\lambda is the wavelength

d is the separation between the slits

In this problem, the separation between the slits decreases: this means that d in the formula decreases. As we see, the value of sin \theta (and so, also \theta) is inversely proportional to d: so, if the d decreases, then the angular separation between the fringes increases.

So, the correct answer is

The distance between interference fringes increases.

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Answer:

A. Technician A only.B.

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Nitrogen at low pressure has also been used to check for leakage in the fuel system of a vehicle.

4 0
3 years ago
Which element can combine with chlorine to make the best conductor of electricity when in the liquid form?
Svetach [21]
The best and most correct answer among the choices provided by your question is the third choice or letter C.

Lithium is the <span>element that can combine with chlorine to make the best conductor of electricity when in the liquid form because </span><span>it has a much lower electronegativity value.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
3 0
2 years ago
The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg astronaut inside the station pushes off one wall of the st
Aleonysh [2.5K]

Answer:

a = 5.83 \times 10^{-4} m/s^2

Explanation:

Since the system is in international space station

so here we can say that net force on the system is zero here

so Force by the astronaut on the space station = Force due to space station on boy

so here we know that

mass of boy = 70 kg

acceleration of boy = 1.50 m/s^2

now we know that

F = ma

F = 70(1.50) = 105 N

now for the space station will be same as above force

F = ma

105 = 1.8 \times 10^5 (a)

a = \frac{105}{1.8 \times 10^5}

a = 5.83 \times 10^{-4} m/s^2

3 0
3 years ago
During combustion, a fuel’s electromagnetic energy is converted into thermal energy.
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6 0
3 years ago
A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of
Nastasia [14]

Answer:

Amplitude = 0.058m

Frequency = 6.25Hz

Explanation:

Given

Amplitude (A) = 8.26 x 10-2 m

Frequency (f) = 4.42Hz

Conversation of energy before split

½mv² = ½KA²

Make A the subject of formula

A = v\sqrt{\frac{m}{k} }

Conversation of energy after split

½(m/2)V'² = ½(m/2)V² = ½KA'²

½(m/2)V² = ½KA'²

Make A the subject of formula

First divide both sides by ½

(m/2)V² = KA'²

Divide both sides by K

\frac{m}{2K}V² = A'²

v\sqrt{\frac{m}{2k} } = A'

Substitute v\sqrt{\frac{m}{k} } for A in the above equation

A' = A/√2

A' = 8.26 x 10^-2/√2

A' = 0.05840702012600882

Amplitude after split = 0.058 (Approximated)

Frequency (f') = f√2

f' = 4.42√2

f' = 6.25082394568908011

Frequency after split = 6.25Hz (approximated)

8 0
3 years ago
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