Bending at 40° — phryzby

Q. Light in air meets water at 40° to the normal. What is the angle of refraction? expects A

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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?

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The solution

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 Ruby

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

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

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

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?