A piece of glass with a flat surface is at the bottom of a tank of water. If a ray of light traveling in the glass is incident o
1 answer:
Answer:
λ₀ = 542.6 nm
Explanation:
The speed of light is related to the wavelength and the frequency by the equation
c = λ f
When a beam of light strikes a surface, it makes the electrons in it oscillate, this is a forced oscillation, so the oscillation frequency equals the frequency of the intendent radiation, then the electrons radiate the radiation for a while that absorbed; therefore, the light in both media has the same frequency
As the speed of light changes in each medium, the only way to comply with the equation is for the wavelength of the radiation to change, let's use the definition of the index of refraction
v = f
n = c / v
n = λ₀ f / f
n = λ₀ /
= λ₀ / n
Let's apply this equation to our case
In the glass λ = 408 nm
In the water is
= λ₀ / n
λ₀ =n
The index of refraction of water-related glass can be obtained by the law of refraction
Water glass
n₁ sin θ₁ = n₂ sin θ₂
For an incident angle of 50º the beam is not refracted in the water, this means that it is parallel to the surface of the glass therefore the angle is 90º, this is called internal total reflection
for angles less than 50º the light that refracts the water that has refractive index 1.33 changes its wavelength
calculate
= λ₀ / n
λ₀ = n
λ₀ = 1.33 408
λ₀ = 542.6 nm
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Answer:
154° at 195 km/h
Explanation:
The helicopter is moving south at 175 km/h, relative to the wind.
But the wind is moving east at 85 km/h, relative to the ground.
This means that the helicopter is moving south east relative to the ground.
Every hour, the helicopter will move 175 km to the south and 85 km to the east, relative to the ground.
This means that we can determine the speed and direction of the helicopter using a right triangle and simple trigonometry.
Refer to the triangle b1.
The distance traveled by the helicopter in 1 hour is denoted by d.
d is the hypotenuse of the right triangle.
Using the Pythagorean Theorem, we can calculate d to be 195 km (rounded to 3 s. f.)
Hence the helicopter is traveling at 195 km/h relative to the ground.
To calculate the direction we use,
tan (x) = opposite/adjacent = 85/175
So the angle x is,
= 25.9°
Relative to the North, the helicopter is moving at 180° - 25.9° = 154° (rounded to 3 s. f.)
A)
B)
C)
D)
E)
Explanation:
A)
When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:
where
q is the charge of the particle (positive)
On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):
where
m is the mass of the particle
v is its final speed
According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:
And solving for v, we find the speed v at which the particle enters the cyclotron:
B)
When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is
where B is the strength of the magnetic field.
This force acts as centripetal force, so we can write:
where r is the radius of the orbit.
Since the two forces are equal, we can equate them:
And solving for r, we find the radius of the orbit:
(1)
C)
The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.
It can be calculated as the ratio between the length of the circumference () and the velocity of the particle (v):
(2)
From eq.(1), we can rewrite the velocity of the particle as
Substituting into(2), we can rewrite the period of revolution of the particle as:
And we see that this period is indepedent on the velocity.
D)
The angular frequency of a particle in circular motion is related to the period by the formula
(3)
where T is the period.
The period has been found in part C:
Therefore, substituting into (3), we find an expression for the angular frequency of motion:
And we see that also the angular frequency does not depend on the velocity.
E)
For this part, we use again the relationship found in part B:
which can be rewritten as
(4)
The kinetic energy of the particle is written as
So, from this we can find another expression for the velocity:
And substitutin into (4), we find:
So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.
Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).