Displacement from the center line for minimum intensity is 1.35 mm , width of the slit is 0.75 so Wavelength of the light is 506.25.
<h3>How to find Wavelength of the light?</h3>
When a wave is bent by an obstruction whose dimensions are similar to the wavelength, diffraction is observed. We can disregard the effects of extremes because the Fraunhofer diffraction is the most straightforward scenario and the obstacle is a long, narrow slit.
This is a straightforward situation in which we can apply the
Fraunhofer single slit diffraction equation:
y = mλD/a
Where:
y = Displacement from the center line for minimum intensity = 1.35 mm
λ = wavelength of the light.
D = distance
a = width of the slit = 0.75
m = order number = 1
Solving for λ
λ = y + a/ mD
Changing the information that the issue has provided:
λ = 1.35 * 10^-3 + 0.75 * 10^-3 / 1*2
=5.0625 *10^-7 = 506.25
so
Wavelength of the light 506.25.
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Answer:
5.23km/s
Explanation:
Given
Radius of Earth = 6.37 * 10^6 m
Altitude of Satellite = 8200km = 8200 * 10³m = 8.2 * 10^6 m
Gravity Acceleration on Satellite Altitude = 1.87965m/s²
For a satellite to remain in circular orbit, then it means the acceleration of gravity must be exact as the centripetal acceleration.
Centripetal Acceleration = V²/R
So, Acceleration of Gravity (A)= Centripetal Acceleration = V²/R
Make V the subject of formula
A = V²/R
V² = AR
V = √AR
Where R = (radius of earth) + (altitude of satellite)
R = 6.37 * 10^6 + 8.2 * 10^6
R = 14.57 * 10^6m
A = 1.87965m/s²
V = √(1.87965 * 14.57x10^6)
V = √27386500.5
V = 5233.211299001789
V = 5233.2113 m/s ------- Approximated
V = 5.23km/s
The answer is 4,045.1 meters
Answer:
A
Explanation:
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