To solve the problem it is necessary to apply the Malus Law. Malus's law indicates that the intensity of a linearly polarized beam of light, which passes through a perfect analyzer with a vertical optical axis is equivalent to:

$I=I_0 cos^2\theta$

Where,

$I_ {0}$ indicates the intensity of the light before passing through the polarizer,

I is the resulting intensity, and

$\theta$ indicates the angle between the axis of the analyzer and the polarization axis of the incident light.

Since we have two objects the law would be,

$I=I_0cos^2\theta_1*cos^2(\theta_2-\theta_1)$

Replacing the values,

$I=100*cos^2(20)*cos^2(40-20)$

$I=100*cos^4(20)$

$I=77.91W/m^2$

Therefore the intesity of the light after it has passes through both polarizers is $77.91W/m^2$

7 0

3 years ago

HELP WILL MARK BRAINLIEST!!

kherson [118]

Answer:15

Explanation:

8 0

3 years ago

A 6.72 particle moves through a region of space where an electric field of magnitude 1270 N/C points in the positive direction,

Romashka [77]

Answer:

$v_{y} = -104 m/s$

Explanation:

Using:

Force = electric field * charge

$F=e*q$

Force = magnitude of charge * velocity * magnetic field * sin tither

$F_{x2}= |q|*v*B*sin \alpha$

Force on particle due to electric field:

$F_{x1}= E*q = (1270N/C)*(-6.72*10^{-6} ) = -8.53*10^{-3}$

Force on particle due to magnetic field:

$F_{x2}= |q|*v*B*sin \alpha = (6.72*10^{-6} )*(1.15)*(sin90)*v = (7.728*10^{-6})*(v)$

$F_{x2}$ is in the positive x direction as $F_{x1}$ is in the negative x direction while net force is in the positive x direction.

Magnetic field is in the positive Z direction, net force is in the positive x direction.

According to right hand rule, Force acting on particle is perpendicular to the direction of magnetic field and velocity of particle. This would mean the force is along the y-axis. As this is a negatively charged particle, the direction of the velocity of the particle is reversed. Therefore velocity of particle, v, has to be in the negative y direction.

Now,

$F_{xnet}- F_{x1 } = F_{x2 }$