Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that it
1 answer:
Answer:
a) I= I₀ (cos²θ - cos⁴θ) b) 75.5º
Explanation:
a) For this exercise we must use Malus's law
I = I₀ cos² θ
where tea is the angle between the two polarizers.
We apply this expression to our case
* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical
I₁ = I₀ cos² θ
* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.
The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical
θ₂ = 90- θ’ = θ
fiate that in the exercise we must take a reference system and measure everything with respect to this system.
I = I₁ cos² θ'
we substitute
I = (I₀ cos² tea) cos² (θ - 90)
cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ
I = Io cos² θ sin² θ
1= cos²θ+ sin²θ
sin²θ = 1 - cos²θ
I= I₀ (cos²θ - cos⁴θ)
b) to find when the intensity is maximum,
we can use that we have an extreme point when the drift is zero
= 0
\frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0
whereby
cos θ - 2 cos³ θ = 0
cos θ ( 1 - 2 cos² θ) = 0
The zeros of this function are in
θ = 90º
1-2cos²θ =0 cos θ = 0.25 θ = 75.5º
Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.
Consequently, the angle that allows the maximum intensity to pass is 75.5º
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Answer:
the magnetic field experienced by the electron is 0.0511 T
Explanation:
Given the data in the question;
Wavelength λ = 21 cm = 0.21 m
we know that Bohr magneton μ is 9.27 × 10⁻²⁴ J/T
Plank's constant h is 6.626 × 10⁻³⁴ J.s
speed of light c = 3 × 10⁸ m/s
protein spin causes magnetic field in hydrogen atom.
so
Initial potential energy = -μB × cos0°
= -μB × 1
= -μB
Final potential energy = -μB × cos180°
= -μB × -1
= μB
so change in energy will be;
ΔE = μB - ( -μ
B )
ΔE = 2μB
now, difference in energy levels will be;
ΔE = hc/λ
2μB = hc/λ
2μBλ = hc
B = hc / 2μλ
so we substitute
B = [(6.626 × 10⁻³⁴) × (3 × 10⁸)] / [2(9.27 × 10⁻²⁴) × 0.21 ]
B = [ 1.9878 × 10⁻²⁵ ] / [ 3.8934 × 10⁻²⁴ ]
B = 510556326.09
B = 0.0511 T
Therefore, the magnetic field experienced by the electron is 0.0511 T
The time period of the pendulum is affected by the acceleration due to gravity. The tension does not have any effect on the time period of the pendulum and also the mass of the bob does not effect the time period of the pendulum.
Hence, if the gravity increases then the time period of the pendulum will decreases and it will swing faster.
Answer:
(a) 28690 m/s (b)
Explanation:
The orbital speed is define as:
(1)
Where r is the radius of the trajectory and T is the orbital period.
To determine the orbital speed of the Earth it is necessary to know the orbital period and the radius of the trajectory. That can be done by means of the Kepler's third law and average velocity equation.
The average velocity in a Uniform Rectilinear Motion is defined as:
(2)
Where v is the velocity, d is the covered distance and t is the time.
Equation 2 can be rewritten for d to get:
(3)
In this case, v will be the speed of light and t, the 8 minutes that takes to reach the Earth.
The time will be converted to seconds so the units in equation 3 can match:
⇒
Replacing all those values in equation 3 it is gotten:
Kepler’s third law is defined as:
Where T is orbital period and r is the radius of the trajectory.
It is necessary to pass from meters to astronomical unit (AU), 1 AU is defined as the distance between the Earth and the Sun.
That can be expressed in units of years.
But there are 31536000 seconds in one year:
Finally, equation 1 can be used:
<u>So Earth orbital speed around the Sun is 28690 m/s.</u>
<em>b) What is its kinetic energy?</em>
The kinetic energy is defined as:
(4)
Notice that it is necessary to found the mass of the Earth, that can be done combining the Universal law of gravity and Newton's second law:
(5)
M will be isolated in equation 5:
Where r is the radius of the Earth ()
<u>Hence, the kinetic energy of Earth is .</u>