A laser produces light that is polarized in the vertical...

A laser produces light that is polarized in the vertical direction. The light travels in the positive $y$ -direction and passes through two polarizers, which have polarizing angles of $35.0^{\circ}$ and $55.0^{\circ}$ from the vertical, as shown in the figure. The laser beam is collimated (neither converging nor expanding), has a circular cross section with a diameter of $1.00 \mathrm{~mm}$, and has an average power of $15.0 \mathrm{~mW}$ at point $A$. At point $C$, what are the amplitudes of the electric and magnetic fields, and what is the intensity of the laser light?

Key Concepts

Polarization

Polarization describes the orientation of the electric field vector of a light wave. In this context, understanding polarization is critical because the behavior of light as it passes through polarizers depends on the alignment between the light's polarization direction and the polarizer's transmission axis.

Malus's Law

Malus's Law is a key concept that describes how the intensity of polarized light is reduced when it passes through a polarizing filter. The transmitted intensity is proportional to the cosine squared of the angle between the incoming light's polarization direction and the axis of the polarizer, which influences the amplitude of the light that emerges.

Electromagnetic Wave Field Amplitudes

The electric and magnetic field amplitudes are central in describing an electromagnetic wave. Their magnitudes determine the energy and intensity of the wave. Calculations relating the power of the beam to the fields involve understanding how a change in amplitude through optical elements (such as polarizers) affects the overall intensity of the light.

Intensity and Power Distribution

Intensity is defined as the power per unit area and is directly related to the square of the electromagnetic field amplitudes. In problems involving laser beams with a known power and cross-sectional area, converting these quantities into field amplitudes is a crucial step in analyzing how the beam interacts with various optical components.

Transcript

00:01 Question we have to calculate the amplitude of electric field and magnetic field and intensity of the light okay these three values we have to calculate so first of all calculating intensity of light after passing through the series of two polarizers so intensity i after first polarizer it is i1 equals to i not which is initial intensity okay so i not cos square theta 1 okay where theta 1 is equal to theta 1 is equal to 55 degree and the intensity after second polarizer it will be given by i 1 cos square 90 uh theta 2 minus theta 1 okay theta 2 minus theta 1 okay so from after combining these two equations we can write that i 2 it will be equals to i not cos square theta 1 and cos square theta 2 minus theta 1 okay now intensity i not this can be written as in the term of power so i not it will be equals to p by area and area will be equals to area can be written as pi by 4 multiplied by diameter d square so from here we will get 4 p divided by d square so substituting value of i not here so we will get that intensity i 2 it will be equals to 4 p divided by d square and and cos square theta 1, cos square theta 2 minus theta 1.

01:29 So now substituting values, so we get 4 multiplied by power which is 15 millie watts, so 15 .0 into 10 to the power minus 3 watt divided by diameter d which is pi also, okay? so pi multiplied by d which is 1 .00 multiplied by 10 to the power minus 3 whole square and cos square theta 1 which is 35 degree and cos square theta 2 which is 55 degree minus theta 1 which is 35 degree so from here after solving we will get that intensity i 2 it is equals to 1 .13 mudplied by 10 to the power 4 watt per meter square okay so this is the answer for the intensity that is passed from the series of two polarizer okay now we can calculate the electric field e so it will be given by it will be given by the relation i intensity which is equals to 1 by 2 c multiplied by mu not and multiplied by amplitude of electric field square okay so from here after rearranging electric field e comes out to be 2 i c multiplied by mu not and whole under root so substituting values where this intensity i it is equals to this intensity i to so, e, this will be equals to under root of 2 multiplied by i, which is 1 .13 multiplied by 10 to the power 4, multiplied by c, which is speed of light 3 into 10 to the power 8, multiplied by mu 0, 4 pi multiplied by 10 to the power minus 7.

03:08 So from here after solving, amplitude of electric field, e, this will be equals to 2 .92 multiplied by 10 to the power 3 volt per meter...

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