Light of original intensity I0 passes through two ideal...

Light of original intensity $I_{0}$ passes through two ideal polarizing filters having their polarizing axes oriented as shown in Fig. E33.30. You want to adjust the angle $\phi$ so that the intensity at point $P$ is equal to $I_{0} / 18$. (a) If the original light is unpolarized, what should $\phi$ be? (b) If the original light is linearly polarized in the same direction as the polarizing axis of the first polarizer the light reaches, what should $\phi$ be?

Key Concepts

Intensity Reduction through Sequential Polarizers

When light passes through a sequence of polarizing filters, each filter's effect depends on its angle relative to the light's polarization direction. For unpolarized light, the first polarizer reduces the intensity by half, after which Malus's Law applies for any additional polarizers. This concept is essential for analyzing how multiple polarizing elements in series can control and reduce light intensity in optical setups.

Malus's Law

Malus's Law quantifies the intensity of polarized light transmitted through a polarizing filter. It states that the transmitted intensity is proportional to the square of the cosine of the angle between the light's initial polarization direction and the axis of the filter. This principle is key for calculating intensity changes when light passes through one or more polarizers.

Polarization

Polarization describes the directional orientation of the oscillations of the electromagnetic fields in a light wave. Light can be unpolarized, meaning its waves oscillate in all directions equally, or it can be linearly polarized, where the oscillations occur along a specific direction. This concept is fundamental in understanding how light interacts with optical elements like polarizers.

Polarizing Filters

Polarizing filters are optical devices that allow light oscillating in one particular direction to pass through while blocking light oscillating in other directions. They are crucial in controlling the state of polarization of light and are used in various applications including photography, displays, and optical experiments.

Transcript

00:01 Hi there, so for this problem, we're told that light of original intensity isu0 passes through two ideal polarizing filters, having their polarizing axis orientated as the figure that is shown in the book.

00:17 Now, you want to adjust the angle fee, so that intensity at a point b is equal to initial intensity divided by 18.

00:26 So what we are going to set for this problem is that intensity is equal to to the initial intensity divided by 18, where i is the intensity of the light passed by the second polarizer.

00:42 Now, for part a of this problem, the question is, if the original light is unpolarance, what should be the angle fee? now, first of all, we know that when unpolarized light passes through a polarizer, the intensity is reduced by a factor of one, divided by 2.

01:05 And the transmitted light is polarized along the axis of the polarizer.

01:10 Okay? so, in this case, after the first filter, we know that the intensity is now the initial intensity divided by 2.

01:21 And the light is polarized along the vertical direction.

01:25 Now, after the second filter, we will want to set the intensity now to the initial intensity divided by 18.

01:33 So what we are going to have now is the initial intensity divided by 18.

01:38 Then this is equal to the initial intensity divided by 2, and this times the cosine of fee and that to the square.

01:48 So now we can solve for the then the cosine of fee.

01:58 In this case, is just simply equal to the square root of 1 divided by 9...

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