This is an excellent question that i do not have the answer to.
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:
They experience the same magnitude impulse
Explanation:
We have a ping-pong ball colliding with a stationary bowling ball. According to the law of conservation of momentum, we have that the total momentum before and after the collision must be conserved:
where is the initial momentum of the ping-poll ball
is the initial momentum of the bowling ball (which is zero, since the ball is stationary)
is the final momentum of the ping-poll ball
is the final momentum of the bowling ball
We can re-arrange the equation as follows or
which means (1) so the magnitude of the change in momentum of the ping-pong ball is equal to the magnitude of the change in momentum of the bowling ball.
However, we also know that the magnitude of the impulse on an object is equal to the change of momentum of the object:
(2) therefore, (1)+(2) tells us that the ping-pong ball and the bowling ball experiences the same magnitude impulse: