A Polaroid Between Crossed Polaroids

Q. A polaroid sheet is placed between two crossed polaroids, with its pass-axis at 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? expects 0.25

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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?

The playground

Inputs

angle

θ_{m}

45.00°

Outputs

t_{b}

The solution

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 Ruby

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

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

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

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