Three complete orders on each side of the m=0 order can be produced in addition to the m=0 order.
The ruling separation is d=1/(470mm-1)
Diffraction lines occurs at an angle θ such that dsin=mλ,when λ is the wavelength and m is an integer.
Notice that for a given order,the line associated with a long wavelength is produced at a greater angle than the line associated with shorter wavelength.
we take λ to be the longest wavelength in the visible spectrum (538nm) and find the greatest integer value of m such that θ is less than 90°.
That is,find the greater integer value of m for which mλ<d.
since,d/λ
There are three complete orders on each side of the m=0 order.
The second and third orders overlap.
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Pushing a broke down car, even done by more than one person, is difficult especially if the distance to be covered is quite far. A car is heavy and it requires a lot of force to start the car moving. This is because the inertia of the car to remain at rest is great. Additionally, the force applied in pushing the car must be greater than the frictional force to cause it to accelerate. The frictional force is dependent on the mass of the object which means that the frictional force acting on the car is also great. Finally, with every push of the car, the frictional force will always be present and acting on the opposite direction. The push that will be supplied must be sustained all throughout.
0.23 mm far apart are the second-order fringes for these two wavelengths on a screen 1.5 m away.
<h3>Given wavelengths 710nm and 660nm,0.65mm apart two slits, and a screen 1.5m away.</h3>
Position of n the order fringe = n λ D / d
for n = 2
position = 2 λ D / d
λ = 710 nm , D = 1.5m
d = .65 x 10⁻³
position 1 = 2 x 710 x 10⁻⁹ x 1.5 / .65 x 10⁻³
= 3276.92 x 10⁻⁶ m
= 3.276x 10⁻³ m
= 3.276mm .
For λ = 660 nm
position = 2 λ D / d
λ = 660 nm , D = 1.5 m
d = .65 x 10⁻³
position 2 = 2 x 660 x 10⁻⁹ x 1.5 / .65 x 10⁻³
= 3046.15 x 10⁻⁶ m
= 3.046 x 10⁻³ m
= 3.046 mm .
Difference between their position
= 3.276mm ₋ 3.046 mm
= 0.23 mm .
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Answer: The force does not change.
Explanation:
The force between two charges q₁ and q₂ is:
F = k*(q₁*q₂)/r^2
where:
k is a constant.
r is the distance between the charges.
Now, if we increase the charge of each particle two times, then the new charges will be: 2*q₁ and 2*q₂.
If we also increase the distance between the charges two times, the new distance will be 2*r
Then the new force between them is:
F = k*(2*q₁*2*q₂)/(2*r)^2 = k*(4*q₁*q₂)/(4*r^2) = (4/4)*k*(q₁*q₂)/r^2 = k*(q₁*q₂)/r^2
This is exactly the same as we had at the beginning, then we can conclude that if we increase each of the charges two times and the distance between the charges two times, the force between the charges does not change.
