Bending at 40° — phryzby

solved Refraction

Q. Light in air $μ_{a} = 1$ meets water $μ_{w} = 1.33$ at 40° $θ_{l} (i) = 40 °$ to the normal. What is the angle of refraction?

How it's solved

Solved with Refraction

Given

$θ_{l} (i) = 40 °$
$μ_{a} = 1$ from “air”
$μ_{w} = 1.33$ from “water”

Find $θ_{l} (r)$ .

Working

snells law(incident, refracted) $μ_{a} \cdot \sin θ_{l} (i) = μ_{w} \cdot \sin θ_{l} (r)$
$1 \cdot \sin 40 ° = 1.33 \cdot \sin θ_{l} (r)$
$θ_{l} (r) = 28.90°$

Conclusion

Going into the denser medium the ray bends toward the normal, so it comes out at a smaller angle than it went in at.

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 :bending_at_forty do4  kind :scq5  describes "Light in air meets water at %{angle} to the normal. What is the angle of refraction?"6  given angle: "40 deg"78  option 28.9.deg9  option 40.0.deg10  option 58.77.deg11  option 50.0.deg12end

solution.rb

1require_relative "../questions/bending_at_forty"2require_relative "../science/ray_optics_and_optical_instruments/refraction"3require_relative "../science/ray_optics_and_optical_instruments/refractive_media"45solve :bending_at_forty, using: Refraction do6  objects :light, :air, :water7  light.goes(:incident, through: air)8  light.goes(:refracted, through: water, after: incident)9  take angle: light.angle_to_the_normal(during: incident)10  deduce_from "air", air.refractive_index => REFRACTIVE_MEDIA.fetch("air")11  deduce_from "water", water.refractive_index => REFRACTIVE_MEDIA.fetch("water")12  answer find_value(light.angle_to_the_normal(during: refracted), in: "deg")1314  conclusion "Going into the denser medium the ray bends toward the normal, so it comes out at a " \15             "smaller angle than it went in at."16end

playground.rb

1require_relative "../solutions/bending_at_forty"2require_relative "../pictures/light"34play_with :bending_at_forty do5  title "Bending at 40°"67  draw Light::Picture do8    media "air", "water"9    ray "incident", arriving_at: :i, angle: true, extended: true10  end11end

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
refractive index of air	$μ_{a}$	—	—
angle to the normal of light during incident	$θ_{l} (i)$	—	0 … 1.571
refractive index of water	$μ_{w}$	—	—
angle to the normal of light during refracted	$θ_{l} (r)$	—	0 … 1.571
refractive index of light	$μ_{l}$	—	—

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_{a} \cdot \sin θ_{l} (i) = μ_{w} \cdot \sin θ_{l} (r)

snells law(incident, refracted)

μ_{a} \cdot \sin c = μ_{w} \cdot \sin 90 °

snells law when grazing exit(incident, refracted)

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?