Ask question

Ask question

All categories

3 years ago

9

A polarized light that has an intensity I0 = 60.0 W/m² is incident on three polarizing disks whose planes are parallel and cente

red on a common axis. Suppose that the transmission axis of the first polarizer is rotated 20° relative to the axis of polarization of the incident light, and that the transmission axis of each additional analyzer is rotated 20° relative to the transmission axis of the previous one. Calculate the transmitted intensity through all polarizers.

1 answer:

nikitadnepr [17]3 years ago

6 0

Answer:

The transmitted intensity through all polarizers is $I_3 =41.31 W/m^2$

Explanation:

According to Malu's law the intensity of a polarized light having an initial intensity $I_0$ is mathematically represented as

$I = I_0cos^2 \theta$

Now considering the polarizer(The polarizing disk) the equation above becomes

$I = I_0 (cos^2 \theta)^n$

Where n is the number of polarizers

Substituting $60.0W/m^2$ for the initial intensity 3 for the n and 20° for the angle of rotation

$I_3 = 60 (cos^220)^3$

$=41.31 W/m^2$

You might be interested in

Displacement vectors of 4km north, 2km south, 5km north, 5km south combine to a total displacement of

goldfiish [28.3K]

<u>Answer</u>

The combined displacement is 2km north

<u>Explanation</u>

Since displacement is a vector quantity, we take into account the direction.

Good for us all the displacement vectors are in the same dimension, so we can make north positive and south negative or vice-versa.

We now add to obtain,

$4+-2+5+-5$

This will simplify to

$=4-2+5-5=2$

Therefore the combined displacement is 2km north

5 0

3 years ago

Select the correct answer.

Alenkasestr [34]

The answer would be c because it is talking about she wants to be in a good neighborhood

3 0

3 years ago

Read 2 more answers

A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turn

kumpel [21]

Answer:

a) The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

Explanation:

Statement is incomplete. The complete description is now described below:

<em>A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turns on, causing an acceleration of 0.250 m/s2 in the x direction. The acceleration lasts for 45.0 s, at which point the thruster turns off. </em>

<em>(a) What is the magnitude of the satellite's velocity when the thruster turns off</em>

<em>(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis. ° counterclockwise from the +x-axis</em>

Let be x and y-directions orthogonal to each other and the satellite is accelerated uniformly from rest in the +x direction and moves at constant velocity in the +y direction. The velocity vector of the satellite ( $\vec{v}_{S}$ ), measured in meters per second, is:

$\vec{v}_{S} = (v_{o,x}+a_{x}\cdot t)\,\hat{i}+v_{y}\,\hat{j}$

Where:

$v_{o,x}$ - Initial velocity in +x direction, measured in meters per second.

$a_{x}$ - Acceleration in +x direction, measured in meter per square second.

$t$ - Time, measured in seconds.

$v_{y}$ - Velocity in +y direction, measured in meters per second.

If we know that $v_{o,x} = 0\,\frac{m}{s}$ , $a_{x} = 0.250\,\frac{m}{s^{2}}$ , $t = 45\,s$ and $v_{y} = 21.4\,\frac{m}{s}$ , the final velocity of the satellite is:

$\vec{v}_{S} = \left[0\,\frac{m}{s}+\left(0.250\,\frac{m}{s^{2}} \right)\cdot (45\,s) \right]\,\hat{i}+\left(21.4\,\frac{m}{s} \right)\,\hat{j}$

$\vec{v_{S}} = 11.25\,\hat{i}+21.4\,\hat{j}\,\,\left[\frac{m}{s} \right]$

a) The magnitud of the satellite's velocity can be found by the resource of the Pythagorean Theorem:

$\|\vec {v}_{S}\| = \sqrt{\left(11.25\,\frac{m}{s} \right)^{2}+\left(21.4\,\frac{m}{s} \right)^{2}}$

$\|\vec{v}_{S}\| \approx 24.177\,\frac{m}{s}$

The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is determined with the help of trigonometric functions:

$\tan \alpha = \frac{v_{y}}{v_{x}} = \frac{21.4\,\frac{m}{s} }{11.25\,\frac{m}{s} }$

$\tan \alpha = 1.902$

$\alpha = \tan^{-1}1.902$

$\alpha \approx 62.266^{\circ}$

The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

4 0

3 years ago

What is 60mph (miles per hour) in meters per second? ( A mile is 5280ft)<br> please someone help me

777dan777 [17]

Answer:

60mph=26.8224meters per second

Explanation:

4 0

3 years ago

A rock is dropped from a vertical cliff. The rock takes 6.00 s to reach the ground below the cliff. What is the height of the cl

Lana71 [14]

Answer:

180 m

Explanation:

The rock follows a free-fall motion - so the vertical distance covered can be found by using the equation

$h=\frac{1}{2}gt^2$

where

g = 10 m/s^2 is the acceleration due to gravity

t = 6.00 s is the time of the fall

Substituting these data, we find the height of the cliff:

$h=\frac{1}{2}(10 m/s^2)(6.00 s)^2=180 m$

4 0

3 years ago