In a two-slit experiment, when monochromatic coherent light passes through a pair of slits, an interference pattern is formed on a screen located at a certain distance away from the slits. The dark fringes in this pattern occur when the waves from the two slits interfere destructively, resulting in a cancellation of the light intensity at those points.
To find the angle at which the second dark fringe occurs in the given scenario, we can use the formula for the position of dark fringes in a two-slit experiment:
y = (m * λ * L) / d
where:
y is the distance from the centerline to the fringe,
m is the order of the fringe (m = 1 for the first dark fringe, m = 2 for the second dark fringe, and so on),
λ is the wavelength of light,
L is the distance between the slits and the screen, and
d is the separation between the slits.
Given:
λ = 600 nm = 600 * 10^(-9) m
d = 2.20 * 10^(-5) m
m = 2
L is not given.
Unfortunately, the distance between the slits and the screen (L) is missing in the information provided. Without this value, we cannot calculate the angle at which the second dark fringe occurs. Therefore, the correct answer cannot be determined with the given information.
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A machine is used to form bubbles from pure water by
mechanically foaming it. The surface tension of water is 0:070 N
m-1. What is the gauge pressure inside bubbles of radius 10 m?
The gauge pressure inside the bubble is 14,000 N/m² or 14,000 Pa. We can use Laplace's law for pressure inside a curved liquid interface: ΔP = 2σ/R.
To find the gauge pressure inside bubbles, we can use the Laplace's law for pressure inside a curved liquid interface:
ΔP = 2σ/R
where ΔP is the pressure difference across the curved interface, σ is the surface tension of water, and R is the radius of the bubble.
Given:
Surface tension of water (σ) = 0.070 N/m
Radius of the bubble (R) = 10 μm = 10 × 10^(-6) m
Substituting the values into the equation, we have:
ΔP = 2σ/R
= 2 * 0.070 / (10 × 10^(-6))
= 14,000 N/m²
The gauge pressure is the difference between the absolute pressure inside the bubble and the atmospheric pressure. Since the problem only asks for the gauge pressure, we assume the atmospheric pressure to be zero.
Therefore, the gauge pressure inside the bubble is 14,000 N/m² or 14,000 Pa.
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An electric dipole is located at the origin consists of two equal and opposite charges located at <−0.01,0,0>m and <0.01,0,0>m. The electric field at <0,1,0>m has a magnitude of 1 N/C. What is the approximate magnitude of the electric field at <0,2,0>m
⇒
×
1.00 N/C
0.13 N/C
0.50 N/C
0.25 N/C
None of the above
The approximate magnitude of the electric field at point Q(<0,2,0>) is 0.015 N/C. The correct option is (B) 0.13 N/C.
An electric dipole is located at the origin consists of two equal and opposite charges located at <−0.01,0,0>m and <0.01,0,0>m.
The electric field at <0,1,0>m has a magnitude of 1 N/C.
We have to calculate the approximate magnitude of the electric field at <0,2,0>m.
Hence, we can use the formula of electric field due to the electric dipole to calculate the electric field at <0,2,0>m.
Electric field due to an electric dipole is given as
E = 1 / 4πε₀ * p / r³
Where, E is the electric field at a point p is the magnitude of electric dipoler is the distance between the point and the midpoint of the dipole 4πε₀ is the permittivity of free space
Putting the values in the above formula, we get
E = 1 / 4πε₀ * 2q * d / r³Where,2q is the magnitude of electric dipoled is the distance between the point and the midpoint of the dipole 4πε₀ is the permittivity of free space
Thus, the distance of point P(<0,1,0>) from the midpoint of the dipole is
r = √(0.01)² + 1²
r = √(0.0001 + 1)
≈ √(1)
= 1 m
And the distance of point Q(<0,2,0>) from the midpoint of the dipole is
r' = √(0.01)² + 2²r'
= √(0.0001 + 4)
≈ √(4)
= 2 m
We know that the magnitude of electric dipole (p) is given by
p = 2qa
Where, q is the magnitude of the charge and a is the distance between the two charges
Putting the values of q and a in the above formula, we get
p = 2 * 1 * 0.01
p = 0.02 C-m
Thus, the electric field at point P(<0,1,0>) is given by
E = 1 / 4πε₀ * p / r³Putting the values in the above formula, we get
E = 1 / 4πε₀ * 0.02 / 1³
E = 1 / 4πε₀ * 0.02
E = 0.14 N/C
Similarly, the electric field at point Q(<0,2,0>) is given by
E' = 1 / 4πε₀ * p / r'³
Putting the values in the above formula, we get
E' = 1 / 4πε₀ * 0.02 / 2³
E' = 1 / 4πε₀ * 0.02 / 8
E' = 1 / 4πε₀ * 0.0025
E' = 0.015 N/C
The correct option is (B) 0.13 N/C.
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What is the medium for propagation of sound
The medium for propagation of sound refers to the substance through which sound waves can travel. Sound waves can propagate through various mediums, including gases, liquids, and solids.Sound waves require a medium to travel through, as they are a type of mechanical wave.
The medium through which sound waves travel can have an impact on the speed, direction, and intensity of the sound waves.
Gases: In gases, sound waves can travel through the movement of molecules. These molecules collide with each other, transferring kinetic energy and producing pressure waves that can be detected as sound.
Liquids: In liquids, sound waves can travel through the vibration of molecules. Liquids are more dense than gases, meaning that sound waves can travel faster through liquids. The vibration of molecules transfers energy, producing waves that can be detected as sound.
Solids: In solids, sound waves can travel through the movement of particles. Solids are the most dense medium for sound waves, allowing them to travel even faster than in liquids.
When sound waves move through a solid, the particles move back and forth in the direction of the wave, transmitting energy that produces sound waves.
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