Unpolarized light and Polarizers

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Unpolarized light passing through two polarizing filters will have its intensity reduced by the first filter, which is oriented at 80 degrees from the y-axis, resulting in an intensity of I0/2. The second filter, oriented at 65 degrees, further modifies this intensity based on the angle difference from the first filter. The correct formula to apply is I2 = I0 * (1/2) * cos^2(θ2 - θ1), where θ2 is 65 degrees and θ1 is 80 degrees. The final intensity of the light after passing through both filters can be calculated using this relationship, ensuring to correctly account for the angle difference. Understanding the effects of each polarizer is crucial for determining the final intensity accurately.
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Unpolarized light propagating along the x-axis from the left in the image above encounters two polarizing filters in a row. The first filter is oriented with its polarization axis at θ1 = 80 degrees from the y-axis, and the second is oriented with its polarization axis at θ2 = 65 degrees from the y-axis. What is the final intensity of the light as a fraction of the original (pre-filter) intensity?

Ive tried doing this problem in every conceivable way possible and I can't get it right.

Please help! I know you have to use a variation of I2=I0*cos^2(theta1) *cos^2(theta2-theta1), but this isn't correct, so what am I doing wrong?

Thanks!
 
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Since the light is initially unpolarized, what is the effect of the first polarizer?
 
Thread 'Variable mass system : water sprayed into a moving container'

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Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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