SME
Grade 11 Science · Problem type
Refraction: Snell’s law
Use this for “Light goes from air into water (n = 1.33) at 40° from the normal. At what angle does it travel in the water, and how fast?”
Diagram
Snell’s law: 1 × sin 40° = 1.33 × sin θ₂
θ₂ = 28.9°: bent toward the normal (into a slower medium).
Sine of the angle out: 0.483
Speed in the second medium: 2.26 × 10⁸ m/s
This is an example. Type your own values to start a new problem.
n₁
sin(
θ₁
°)
=
n₂
sin(
θ₂
°)
s
Sine of the angle out
calculated
v₂
Speed in the second medium
calculated · m/s
Formulas
s
=
n₁
n₂
×
sin(
θ₁
)
0.483
=
1
1.33
×
sin(40°)
θ₂
=
arcsin(
s
)
28.9 = arcsin(0.483)
v₂
=
3 ×
10
8
/
n₂
2.26 ×
10
8
=
3 ×
10
8
/1.33
Assumptions
•
Light slows in glass or water (v = c/n); crossing at an angle, it bends.
•
n₁ sin θ₁ = n₂ sin θ₂, with the angles measured from the normal (dashed).
•
Into a slower medium (bigger n), the ray bends toward the normal.
Step-by-step
WE KNOW
n₁ = 1, n₂ = 1.33, θ₁ = 40°
FIND
sine of the angle out (s), angle out (θ₂), speed in the second medium (v₂)
Step 1 · Find sine of the angle out (s)
Use s = n₁ ÷ n₂ × sin(θ₁)
Snell’
s
law
n₁
sin
θ₁
=
n₂
sin
θ₂
,
solved
for
sin
θ₂
.
s
=
1
1.33
×
sin(40°)
s
=
1
1.33
×
0.6428
s
=
0.7519
×
0.6428
s
=
0.483
Step 2 · Find angle out (θ₂)
Use θ₂ = arcsin(s)
The angle whose sine it is.
θ₂
=
arcsin(0.483)
θ₂
=
28.9°
Step 3 · Find speed in the second medium (v₂)
Use v₂ = 3 × 10⁸/n₂
Light is n times slower in a medium than in a vacuum: c over n.
v₂
=
3 ×
10
8
/1.33
v₂
=
2.26 ×
10
8
m/
s
Check
Put the numbers back in. Both sides should match.
0.483
≈
1
1.33
×
sin(40°)
✓
28.9 = arcsin(0.4833) ✓
2.26 ×
10
8
≈
3 ×
10
8
/1.33
✓
Topic check
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