What is the momentum of an airplane with a mass of 360,000 kg moving down the runway at 1.5 m/s?
1 answer:
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In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
(1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit
In our problem,
while the distance between the first and the fifth minima is
(2)
If we use the formula to rewrite
, eq.(2) becomes
Which we can solve to find a, the width of the slit:
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
D is the distance of the screen from the slit
a is the width of the slit
In our problem,
while the distance between the first and the fifth minima is
If we use the formula to rewrite
Which we can solve to find a, the width of the slit:
Answer:
compared to the incident angle, the refracted angle is 45.56⁰
Explanation:
From Snell's law;
n₁sin(I) = n₂sin(r)
Where;
n₁ is the refractive index of light in medium 1 = 1.2
n₂ is the refractive index of light in medium 2 = 1.4
I is the incident angle
r is the refractive angle
I = 56.439⁰
Applying snell's law
Therefore, compared to the incident angle, the refracted angle is 45.56⁰