How Does Rotating a Polaroid Affect Light Intensity?

[fionamb83]

Two sheets of polaroid are oriented so that there is a maximum transmission of light. One sheet is now rotated by 30 degrees, by what factor does the light intensity drop?



OK, the only equation I could think to use is tanB = N1/N2 but it doesn't seem to work.


the answer is 0.75 I just can't seem to get there.
Any help would be GREATLY appreciated.

 

[JimChampion]

the only equation I could think to use is tanB = N1/N2 but it doesn't seem to work.

Define the terms in this equation: what exactly are B, N1, N2, and how do they relate to the problem you've quoted?

 

[fionamb83]

Got it!

I feel silly! I realize equation I was trying to use was wrong. I solved it using Malus' Law

intensity after = Intensity before*cos^30

Thanks for the quick reply tho.

 

[JimChampion]

Malus' Law is spot on. The intensity is greatest when angle=0deg, zero when angle=90 deg. The cos(angle) function fits the bill, especially as the situation is a rotation.

 

[JimChampion]

Sorry, in the above post I should have said amplitude not intensity.

... Malus' Law

intensity after = Intensity before*cos^30

In the above did you mean $$I = I_0 \sin 30^o$$ or $$I = I_0 \sin^2 30^o$$ ?

 

[fionamb83]

$$I = I_0 \cos^2 30^o$$

From Cutnell, Physics

 

[Antenna Guy]

$$I = I_0 \cos^2 30^o$$

From Cutnell, Physics

$E=E_0 cos(30)$ where $E$ is the field in Volts/m.

Power (intensity) is proportional to $E^2$, which in turn is proportional to $I^2$ (current squared).

$$\frac{I}{I_0}=\frac{E^2}{E_0^2}=\frac{(E_0 cos(30))^2}{E_0^2}=cos^2(30)=0.75$$

[n.b. "I" for intensity is not current - I hope that isn't too confusing]

Regards,

Bill