SME
Grade 11 Science · Problem type
The critical angle
Use this for “What is the critical angle from glass (n = 1.50) into air? Does light hitting the surface at 50° get out?”
Diagram
Snell’s law: 1.5 × sin 50° = 1 × sin θ₂
Critical angle: sin θc = n₂/n₁ = 1/1.5, θc = 41.8°.
Past the critical angle: total internal reflection, no light gets out.
What sin θ₂ would be: 1.15
This is an example. Type your own values to start a new problem.
sin(
θc
°)
=
n₂
n₁
θ₁
Angle in
example · °
s
What sin θ₂ would be
calculated
Formulas
sin(θc)
=
n₂
n₁
sin(41.8°) = 1/1.5
s
=
n₁
n₂
×
sin(
θ₁
)
1.15
=
1.5
1
×
sin(50°)
Assumptions
•
Going into a faster medium (smaller n) the ray bends away from the normal.
•
At the critical angle θc the ray skims the surface: sin θc = n₂/n₁.
•
Past it sin θ₂ would be more than 1: all the light reflects. Optical fibers work this way.
Step-by-step
WE KNOW
n₁ = 1.5, n₂ = 1, θ₁ = 50°
FIND
critical angle (θc), what sin θ₂ would be (s)
Step 1 · Find critical angle (θc)
Use sin(θc) = n₂/n₁
At
the
critical
angle
θ₂
=
90°:
sin
θc
=
n₂
n₁
.
θc
=
arcsin(
n₂
n₁
)
θc = arcsin(1/1.5)
θc = 41.8°
Step 2 · Find what sin θ₂ would be (s)
Use s = n₁ ÷ n₂ × sin(θ₁)
Snell’
s
law
solved
for
sin
θ₂:
more
than
1
means
no
ray
gets
out.
s
=
1.5
1
×
sin(50°)
s
=
1.5
×
0.766
s
=
1.15
(more
than
1,
so
no
ray
gets
out:
all
the
light
reflects)
Check
Put the numbers back in. Both sides should match.
sin(41.8°) = 1/1.5 ✓
1.15
≈
1.5
1
×
sin(50°)
✓
Topic check
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