Question

What angle would the axis of a polarizing filter need to make with the direction of polarized light of intensity $1.00 \textrm{ kW/m}^2$ to reduce the intensity to $10.0 \textrm{ W/m}^2$?

$84.3^\circ$

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Video Transcript

This is College Physics Answers with Shaun Dychko. Polarised light is incident on a polarizing film which then transmits some intensity

*I*and the angle between the polarization axis and the polarization direction of the incident light is what we have to find. And so we have an expression here that the transmitted intensity equals the incident intensity multiplied by a cosine of the angle between the polarizations squared and so we're gonna solve for this*theta*. Because we're given what the transmitted intensity is, 10 watts per square meter, and the incident intensity is one kilowatt per square meter. So, we'll divide both sides by*I naught*and then switch the sides around and we have cos squared*theta*equals*I*over*I naught*and then take the square root of both sides and so we have Cos*theta*equals the square root of the ratios of the intensities and then we'll take the inverse co sine of both sides to get the*theta*the inverse cosine of the square root of the ratio of intensities. So, it’s the inverse cosine of the square root of 10 watts per square meter divided by 1000 watts per square meter, the units here need to be the same by the way and to convert kilowatts into Watts here for the denominator and we end up with 84.3 degrees.