The distance mirror M2 must be moved so that one wavelength has produced one more new maxima than the other wavelength is;
<u><em>L = 57.88 mm</em></u>
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We are given;
Wavelength 1; λ₁ = 589 nm = 589 × 10⁻⁹ m
Wavelength 2; λ₂ = 589.6 nm = 589.6 × 10⁻⁹ m
We are told that L₁ = L₂. Thus, we will adopt L.
Formula for the number of bright fringe shift is;
m = 2L/λ
Thus;
For Wavelength 1;
m₁ = 2L/(589 × 10⁻⁹)
For wavelength 2;
m₂ = 2L/(589.6)
Now, we are told that one wavelength must have produced one more new maxima than the other wavelength. Thus;
m₁ - m₂ = 2
Plugging in the values of m₁ and m₂ gives;
(2L/589) - (2L/589.6) = 2
divide through by 2 to get;
L[(1/589) - (1/589.6)] = 1
L(1.728 × 10⁻⁶) = 1
L = 1/(1.728 × 10⁻⁶)
L = 578790.67 nm
L = 57.88 mm
Read more at; brainly.com/question/17161594
Explanation:
A one-kilogram mass is still a one-kilogram(as mass is an intrinsic property of the object) but the downward force due to gravity, and therefore it's weight, is only one-sixth of what the object would have on the Earth. So man of mass 180 pounds weights only about 30 pounds-force when visiting the moon
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Gravity is an attractive force that works to pull objects together. If 2 objects are close the gravitational pull will be stronger
Mass and distance determine gravity. The farther two things are away from each other, the weaker the gravitational forces are, the less mass an object has the less gravitational force it exerts
The frequency of oscillation is 2.153 Hz
What is the frequency of spring?
Spring Frequency is the natural frequency of spring with a weight at the lower end. Spring is fixed from the upper end and the lower end is free.
For the mass-spring system in this problem,
The Frequency of spring is calculated with the equation:
Where,
f = frequency of spring
k = spring constant = 64 N/m
m = mass attached to spring = 350g = 0.350 kg
a = maximum acceleration = 5.3 m/s^2
Substituting the values in the equation,
Hence,
The frequency of oscillation is 2.153 Hz
Learn more about frequency here:
<u>brainly.com/question/13978015</u>
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