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3 years ago

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An incident electromagnetic wave is polarized and its E vector has a maximum value of 6 Volts/m. It is incident on a polarizer w

hose transmission axis is rotated at an angle of 45 with respect to the incident E vector. What is the maximum value of the transmitted E vector?

1 answer:

adell [148]3 years ago

7 0

To solve this problem it is necessary to apply the equations related to the law of Maus.

By the law of Maus we know that

$I = I_0 cos^2 \theta$

Where,

$I_0$ = Intesity of incident light

I = Intensity of polarized light

With our values we have that

$I_0 =$ 6V/m

$\theta = 45\°$

Then

$I = 6*cos^2(45)$

$I = 3V/m$

Therefore the maximum value of the transmitted E vector is 3V/m

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Answer:

756.88 Volts will be the potential difference across each capacitor.

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Charge of second capacitor = $Q_2$

Voltage across first capacitor = $V_2=560 V$

$Q_2=C_2V_2$

$Q_1=6.61\times 10^{-6} F\times 560 V=0.0037016 C$

Both the capacitors are disconnected and positive plates are now connected to each other and the negative plates are connected to each other. These capacitors are connected in parallel combination.

Total charge = Q

$Q =Q_1+Q_2=0.0034282 C+ 0.0037016 C=0.0071298 C$

Total capacitance in parallel combination:

$C_p=C_1+C_2=2.81\times 10^{-6} F+6.61\times 10^{-6} F=9.42\times 10^{-6} F$

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Answer:

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