Reflectance — phryzby

solved Reflection · Reflectance

Q. Light meets a surface between indices $μ_{1}$ and $μ_{2}$ at $i$ . How much of it gets through?

Inputs

angle of incidence

i

50.00°

refractive index of first medium

μ_{1}

1air

refractive index of second medium

μ_{2}

1.5glass

Outputs

angle of refraction

r_{r}

30.71°

no refracted ray?no

reflectance s polarised

r_{s}

11.2%

reflectance p polarised

r_{p}

0.3%

reflectance

r

5.8%

transmittance

τ

94.2%

How it's solved

Solved with Reflection, Reflectance

Given

$i$ , $μ_{1}$ , $μ_{2}$ yours to vary

Find $r_{r}$ , whether there is no refracted ray, $r_{s}$ , $r_{p}$ , $r$ , $τ$ .

Working

snells law $μ_{1} \cdot \sin i = μ_{2} \cdot \sin r_{r}$
$1 \cdot \sin 50 ° = 1.5 \cdot \sin r_{r}$
$r_{r} = 30.71°$
s polarised $r_{s} = {(\frac{μ_{1} \cdot \cos i - μ_{2} \cdot \cos r_{r}}{μ_{1} \cdot \cos i + μ_{2} \cdot \cos r_{r}})}^{2}$
$r_{s} = {(\frac{1 \cdot \cos 50 ° - 1.5 \cdot \cos 30.71 °}{1 \cdot \cos 50 ° + 1.5 \cdot \cos 30.71 °})}^{2}$
$r_{s} = 11.2%$
p polarised $r_{p} = {(\frac{μ_{1} \cdot \cos r_{r} - μ_{2} \cdot \cos i}{μ_{1} \cdot \cos r_{r} + μ_{2} \cdot \cos i})}^{2}$
$r_{p} = {(\frac{1 \cdot \cos 30.71 ° - 1.5 \cdot \cos 50 °}{1 \cdot \cos 30.71 ° + 1.5 \cdot \cos 50 °})}^{2}$
$r_{p} = 0.3%$
unpolarised $r = \frac{r_{s} + r_{p}}{2}$
$r = \frac{0.112048357 + 0.00327753215}{2}$
$r = 5.8%$
reflected and transmitted share the light $r + τ = 1$
$0.0576629447 + τ = 1$
$τ = 94.2%$
no refracted ray? $i > c$ no

Conclusion

It depends on the angle, and on polarisation: the two polarisations are reflected in quite different measure, and at one angle one of them is not reflected at all.

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 :reflectance do4  describes "Light meets a surface between indices %{first_index} and %{second_index} at %{angle}. %{transmitted}?"5  given :angle6  given :first_index7  given :second_index8  asks transmitted: "How much of it gets through"9end

solution.rb

1require_relative "../questions/reflectance"2require_relative "../science/ray_optics_and_optical_instruments/reflection"3require_relative "../science/ray_optics_and_optical_instruments/reflectance"45solve :reflectance, using: [ Reflection, Reflectance ] do6  take angle: :i, first_index: :mu_1, second_index: :mu_27  find_value :r_r8  find_whether :no_refracted_ray?9  find_value :r_s10  find_value :r_p11  find_value :r12  take transmitted: :tau13  find_value :transmitted1415  conclusion "It depends on the angle, and on polarisation: the two polarisations are reflected in quite different measure, and at one angle one of them is not reflected at all."16end

playground.rb

1require_relative "../solutions/reflectance"2require_relative "../science/ray_optics_and_optical_instruments/refractive_media"3require_relative "../pictures/light"45INDEX = 1.0..4.067play_with :reflectance do8  title "Reflectance"910  input :i, default: 50.deg11  input :mu_1, INDEX, default: 1.0, marks: REFRACTIVE_MEDIA12  input :mu_2, INDEX, default: 1.5, marks: REFRACTIVE_MEDIA1314  output :r_r15  output :no_refracted_ray?, alarm: true16  output :r_s, in: "%"17  output :r_p, in: "%"18  output :r,  in: "%"19  output :tau, in: "%"2021  draw Light::Picture do22    media :mu_1, :mu_223    ray "incident",  arriving_at: :i, angle: true24    ray "reflected", leaving_at: :i, weight: :r25    ray "refracted", crossing_at: :r_r, angle: true,26        weight: :tau, unless: :no_refracted_ray?27    mark "critical", arriving_at: :c28    note "no refracted ray — Fresnel has nothing to say", when: :no_refracted_ray?29  end30end

The laws it drew on

quantity	written	units	within	from
angle to the normal	$θ$	—	0 … 1.571	Incidence
angle of incidence	$i$	—	0 … 1.571	Incidence
angle of reflection	$r_{l}$	—	0 … 1.571	Reflection
refractive index	$μ$	—	—	Refraction
angle of refraction	$r_{r}$	—	0 … 1.571	Refraction
refractive index of first medium	$μ_{1}$	—	—	Refraction
refractive index of second medium	$μ_{2}$	—	—	Refraction
critical angle	$c$	—	0 … 1.571	Refraction
reflectance s polarised	$r_{s}$	—	—	Reflectance
reflectance p polarised	$r_{p}$	—	—	Reflectance
reflectance	$r$	—	—	Reflectance
transmittance	$τ$	—	—	Reflectance

i \equiv θ_{i} (before)

angle of incidence

r_{l} \equiv θ_{i} (after)

angle of reflection

r_{r} \equiv θ_{i} (after)

angle of refraction

μ_{1} \equiv μ_{b}

refractive index of first medium

μ_{2} \equiv μ_{a}

refractive index of second medium

i = r_{l}

law of reflection

μ_{1} \cdot \sin i = μ_{2} \cdot \sin r_{r}

snells law

μ_{1} \cdot \sin c = μ_{2} \cdot \sin 90 °

snells law when grazing exit

r_{s} = {(\frac{μ_{1} \cdot \cos i - μ_{2} \cdot \cos r_{r}}{μ_{1} \cdot \cos i + μ_{2} \cdot \cos r_{r}})}^{2}

s polarised

r_{p} = {(\frac{μ_{1} \cdot \cos r_{r} - μ_{2} \cdot \cos i}{μ_{1} \cdot \cos r_{r} + μ_{2} \cdot \cos i})}^{2}

p polarised

r = \frac{r_{s} + r_{p}}{2}

unpolarised

r + τ = 1

reflected and transmitted share the light

r_{r} = 90 °

grazing exit?

i > c

no refracted ray?

r_{r} < i

bends toward the normal?

r_{r} > i

bends away from the normal?