Out of water at 60°

solved Reflection · Refraction

Q. Light in water $μ_{w} = 1.33$ meets the surface with air $μ_{a} = 1$ at 60° $θ_{l} (i) = 60 °$ to the normal. Which are true?

How it's solved

Solved with Reflection, Refraction

Given

$θ_{l} (i) = 60 °$
$μ_{w} = 1.33$ from “water”
$μ_{a} = 1$ from “air”

Working

snells law when grazing exit(incident, refracted) $μ_{w} \cdot \sin c = μ_{a} \cdot \sin 90 °$
$1.33 \cdot \sin c = 1 \cdot \sin 90 °$
$c = 48.75°$
$θ_{l} (i) > c$ : holds
$θ_{l} (i) < c$ : does not hold
$θ_{l} (i) > c$ : holds
$θ_{l} (r) < θ_{l} (i)$ : cannot be worked out, snells law(incident, refracted) has no solution for the angle to the normal of light during refracted

Conclusion

Past the critical angle Snell's law would need a sine above one, and no angle has that, so there is no refracted ray. All of the light turns back into the water.

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 :out_of_water_at_sixty do4  kind :mcq5  describes "Light in water meets the surface with air at %{angle} to the normal. Which are true?"6  given angle: "60 deg"78  option "No refracted ray leaves the water", as: :none_out9  option "Some of the light gets out into the air", as: :some_out10  option "60° is past water's critical angle", as: :past11  option "The ray bends toward the normal as it leaves", as: :toward12end

solution.rb

1require_relative "../questions/out_of_water_at_sixty"2require_relative "../science/ray_optics_and_optical_instruments/reflection"3require_relative "../science/ray_optics_and_optical_instruments/refraction"4require_relative "../science/ray_optics_and_optical_instruments/refractive_media"56solve :out_of_water_at_sixty, using: [ Reflection, Refraction ] do7  objects :light, :water, :air8  light.goes(:incident, through: water)9  light.goes(:refracted, through: air, after: incident)10  take angle: light.angle_to_the_normal(during: incident)11  deduce_from "water", water.refractive_index => REFRACTIVE_MEDIA.fetch("water")12  deduce_from "air", air.refractive_index => REFRACTIVE_MEDIA.fetch("air")13  take none_out: light.angle_to_the_normal(during: incident) > c,14       some_out: light.angle_to_the_normal(during: incident) < c,15       past: light.angle_to_the_normal(during: incident) > c,16       toward: light.angle_to_the_normal(during: refracted) < light.angle_to_the_normal(during: incident)17  answer options.select { |o| holds?(o.taken) }1819  conclusion "Past the critical angle Snell's law would need a sine above one, and no angle has " \20             "that, so there is no refracted ray. All of the light turns back into the water."21end

playground.rb

1require_relative "../solutions/out_of_water_at_sixty"2require_relative "../pictures/light"34play_with :out_of_water_at_sixty do5  title "Out of water at 60°"67  draw Light::Picture, rising: true do8    media "water", "air"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
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
angle to the normal of light during incident	$θ_{l} (i)$	—	0 … 1.571
refractive index of water	$μ_{w}$	—	—
refractive index of air	$μ_{a}$	—	—
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_{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

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

snells law(incident, refracted)

μ_{w} \cdot \sin c = μ_{a} \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?