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Alexus [3.1K]
2 years ago
10

Linearly polarized light whose Jones vector is [0 1] (horizontally polarized) is sent through a train of two linear polarizers.

The first is oriented with its transmission axis at 45 degrees and the second has its transmission axis vertical. Show that the emerging light is linearly polarized in the vertical direction; that is, the plane of polarization has been rotated by 90 degrees.
Physics
1 answer:
ivann1987 [24]2 years ago
8 0

Answer:

Following are the solution to the given question:

Explanation:

The input linear polarisation was shown at an angle of 2 \mu. It's a very popular use of a half-wave plate. In particular, consider the case \mu = 45 \pm, at which the angle of rotation is 90\pm. HWP thereby provides a great way to turn, for instance, a linear polarised light that swings horizontally to polarise vertically. Illustration of action on event circularly polarized light of the half-wave platform. Customarily it is the slow axis of HWP that corresponds to either the rotation. Note that perhaps the vector of polarization is "double-headed," i.e., the electromagnetic current swinging back and forward in time. Therefore the turning angle could be referred to as the rapid axis to reach the same result. Please find the attached file.

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An incident ray of light strikes a diamond at an angle of
lys-0071 [83]

Answer:

The angle of refraction is option b: 17°.

Explanation:

We can find the angle of refraction by using Snell's law:

n_{1}sin(\theta_{1}) = n_{2}sin(\theta_{2})

Where:

n₁: is the index of refraction of the medium 1 (air) = 1.0003  

n₂: is the index of refraction of the medium 2 (diamond) = 2.42

θ₁: is the angle of incidence = 45°

θ₂: is the angle of refraction =?

Hence, the angle of refraction is:

sin(\theta_{2}) = \frac{n_{1}sin(\theta_{1})}{n_{2}} = \frac{1.0003*sin(45)}{2.42} = 0.2922

\theta_{2} = 17 ^{\circ}

Therefore, the correct option is b: 17°.

I hope it helps you!

4 0
3 years ago
Read 2 more answers
A particle initially located at the origin has an acceleration of = 1.00ĵ m/s2 and an initial velocity of i = 6.00î m/s. (a) Fin
Ira Lisetskai [31]

Answer:

a)     d = (6.00 t i ^ + 0.500 t²) m , b)   v = (6.00 i ^ + 1.00 t j ^) m / s

c) d = (24.00 i ^ + 8.00 j^ ) m , d)  v = (6.00 i ^ + 5 j^ ) m/s

Explanation:

This exercise is about kinematics in two dimensions

a) find the position of the particle on each axis

X axis

Since there is no acceleration on this axis, we can use the relation of uniform motion

       v = x / t

        x = v t

we substitute

        x = 6.00 t

Y Axis

on this axis there is an acceleration and there is no initial speed

         y = v₀ t + ½ a t²

         y = ½ at t²

we substitute

        y = ½ 1.00 t²

        y = 0.500 t²

in vector position is

       d = x i ^ + y j ^

       d = (6.00 t i ^ + 0.500 t²) m

b) x axis

as there is no relate speed is concatenating

       vₓ = v₀

       vₓ = 6.00 m / s

y Axis  

there is an acceleration and the initial speed is zero

         v_{y} = v₀ + a t

         v_{y} = a t

         v_{y} = 1.00 t

the velocity vector is

         v = vₓ i ^ + v_{y} j ^

         v = (6.00 i ^ + 1.00 t j ^) m / s

c) the coordinates for t = 4 s

        d = (6.00 4 i ^ + 0.50 4 2 j⁾

        d = (24.00 i ^ + 8.00 j^ ) m

 

x = 24.0 m

y = 8.00 m

d) the velocity of for t = 4 s

        v = (6 i ^ + 1 5 j ^)

         v = (6.00 i ^ + 5 j^ ) m/s

7 0
3 years ago
3. What does the difference in force depend on?
dangina [55]

Answer:

it depends on the relative masses of the objects.

Explanation:

4 0
3 years ago
Read 2 more answers
What is the MOST LIKELY effect of deforestation that results from urbanization of an area?
KengaRu [80]

A. Erosion is the answer

4 0
3 years ago
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For a maximum superelevation of 0.08 ft/ft and a degree of curve of 4o, calculate the maximum safe speed for the curve assuming
Mamont248 [21]

Answer:

Explanation:

Given that

Superelation= 0.08ft/ft

Given curve= u•

Curve junction factor= 0.13

DR= 5729.57795

R = 5729.57795/D

R = 5729.57795/4

R = 1432.4ft

c + f = V^2/gG

0.08 + 0.13 = V^2 / (32*1432.4)

V^2 = 9625.728 or V = 98 ft/sec

The designed speed for a project considered is a minimum value which means the highway design elements will meet or exceed the standards for the design speed. The maximum safe speed under normal condition is significantly greater than design speed

7 0
3 years ago
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