A Polaroid Between Crossed Polaroids

solved MalusLaw

Q. A polaroid sheet is placed between $t_{b} = t_{m} \cdot t_{l}$ two crossed polaroids $θ_{l} = \frac{π}{2} - θ_{m}$ , with its pass-axis at 45° $θ_{m} = 45 °$ to that of the first. What share of it comes out of the last polaroid of the light that comes out of the first polaroid?

Inputs

angle

θ_{m}

45.00°

Outputs

t_{b}

How it's solved

Solved with MalusLaw

Given

$θ_{m} = 45 °$
$θ_{l} = \frac{π}{2} - θ_{m}$ from “two crossed polaroids”
$t_{b} = t_{m} \cdot t_{l}$ from “placed between”

Find $t_{b}$ .

Working

malus law(middle) $t_{m} = {\cos θ_{m}}^{2}$
$t_{m} = {\cos 45 °}^{2}$
$t_{m} = 0.5$
two crossed polaroids $θ_{l} = \frac{π}{2} - θ_{m}$
$θ_{l} = \frac{3.14159265}{2} - 45 °$
$θ_{l} = 45.00°$
malus law(last) $t_{l} = {\cos θ_{l}}^{2}$
$t_{l} = {\cos 45 °}^{2}$
$t_{l} = 0.5$
placed between $t_{b} = t_{m} \cdot t_{l}$
$t_{b} = 0.5 \cdot 0.5$
$t_{b} = 0.25$
$t_{b} \approx 0.25$ to 2 significant figures

Conclusion

Crossed polaroids alone pass nothing, but a third one between them, at an angle to both, turns the light part of the way each time; at forty-five degrees a quarter gets through.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :a_polaroid_between_crossed_polaroids do4  kind :numerical5  describes "A polaroid sheet is placed between two crossed polaroids, with its pass-axis at %{angle} to that of the " \6            "first. %{share} of the light that comes out of the first polaroid?"7  given angle: "45 deg"8  asks share: "What share of it comes out of the last polaroid"9end

solution.rb

1require_relative "../questions/a_polaroid_between_crossed_polaroids"2require_relative "../science/wave_optics/malus_law"34solve :a_polaroid_between_crossed_polaroids, using: MalusLaw do5  objects :middle, :last, :both67  take angle: middle.angle_to_the_light_s_polarisation89  deduce_from "two crossed polaroids",10              last.angle_to_the_light_s_polarisation => pi / 2 - middle.angle_to_the_light_s_polarisation11  deduce_from "placed between", both.share_passed => middle.share_passed * last.share_passed1213  share = both.share_passed14  answer share:, figures: 21516  conclusion "Crossed polaroids alone pass nothing, but a third one between them, at an angle to both, turns " \17             "the light part of the way each time; at forty-five degrees a quarter gets through."18end

playground.rb

1require_relative "../solutions/a_polaroid_between_crossed_polaroids"23play_with :a_polaroid_between_crossed_polaroids do4  title "A Polaroid Between Crossed Polaroids"5end

The laws it drew on

quantity	written	units	within	from
pi	$π$	—	0 … ∞	MalusLaw
angle to the light s polarisation	$θ$	—	0 … 3.142	MalusLaw
share passed	$t$	—	—	MalusLaw
share passed of ground	$t_{g}$	—	—
angle to the light s polarisation of ground	$θ_{g}$	—	0 … 3.142
share passed of middle	$t_{m}$	—	—
angle to the light s polarisation of middle	$θ_{m}$	—	0 … 3.142
share passed of last	$t_{l}$	—	—
angle to the light s polarisation of last	$θ_{l}$	—	0 … 3.142
share passed of both	$t_{b}$	—	—
angle to the light s polarisation of both	$θ_{b}$	—	0 … 3.142

t_{p} = {\cos θ_{p}}^{2}

malus law

t_{g} = {\cos θ_{g}}^{2}

malus law(ground)

t_{m} = {\cos θ_{m}}^{2}

malus law(middle)

t_{l} = {\cos θ_{l}}^{2}

malus law(last)

t_{b} = {\cos θ_{b}}^{2}

malus law(both)

θ_{l} = \frac{π}{2} - θ_{m}

two crossed polaroids

t_{b} = t_{m} \cdot t_{l}

placed between