Refraction: Air to Water

solved Refraction

Q. Where does a ray go when it passes from air $μ_{1} = 1$ into water $μ_{2} = 1.33$ at $i$ ?

Inputs

angle of incidence

i

60.00°

Outputs

angle of refraction

r_{r}

40.63°

How it's solved

Solved with Refraction

Given

$μ_{1} = 1$ from “air”
$μ_{2} = 1.33$ from “water”
$i$ yours to vary

Find $r_{r}$ .

Working

snells law $μ_{1} \cdot \sin i = μ_{2} \cdot \sin r_{r}$
$1 \cdot \sin 60 ° = 1.33 \cdot \sin r_{r}$
$r_{r} = 40.63°$

Conclusion

It goes through, and bends toward the normal doing so. The two angles are neither equal nor proportional: what each medium does to the sine of its own angle, the other undoes.

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 :refraction do4  describes "%{refracted} when it passes from air into water at %{angle}?"5  given :angle6  asks refracted: "Where does a ray go"7end

solution.rb

1require_relative "../questions/refraction"2require_relative "../science/ray_optics_and_optical_instruments/refraction"3require_relative "../science/ray_optics_and_optical_instruments/refractive_media"45solve :refraction, using: Refraction do6  take angle: :i7  deduce_from "air", mu_1: REFRACTIVE_MEDIA.fetch("air")8  deduce_from "water", mu_2: REFRACTIVE_MEDIA.fetch("water")9  take refracted: :r_r10  find_value :refracted1112  conclusion "It goes through, and bends toward the normal doing so. The two angles are neither equal nor proportional: what each medium does to the sine of its own angle, the other undoes."13end

playground.rb

1require_relative "../solutions/refraction"2require_relative "../science/ray_optics_and_optical_instruments/refractive_media"3require_relative "../pictures/light"45play_with :refraction do6  title "Refraction: Air to Water"78  input :i, default: 60.deg910  output :r_r1112  draw Light::Picture do13    media "air", "water"14    ray "incident",  arriving_at: :i,  angle: true, extended: true15    ray "refracted", crossing_at: :r_r, angle: true16  end17end

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
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

i \equiv θ_{i} (before)

angle of incidence

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

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_{r} = 90 °

grazing exit?

i > c

no refracted ray?

r_{r} < i

bends toward the normal?

r_{r} > i

bends away from the normal?