Out of water at 60°

Q. Light in water meets the surface with air at 60° to the normal. Which are true? expects A, C

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

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?

The playground

The solution

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 Ruby

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

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

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

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?