SME
Electromagnetic Theory · Problem type
Reflection at normal incidence
Use this for “Light passes straight from water (n = 1.33) into diamond (n = 2.42). What fraction of its intensity is reflected, and what fraction goes through?”
Diagram
Snell’s law: 1 × sin 0° = 1.5 × sin θ₂
θ₂ = 0°: bent toward the normal (into a slower medium).
This is an example. Type your own values to start a new problem.
R
=
−
+
2
=
R
T
=
1
−
R
=
T
Formulas
R
=
(
n₁
−
n₂
n₁
+
n₂
)²
0.04
=
(
1 − 1.5
1 + 1.5
)²
T
=
1
−
R
0.96 = 1 − 0.04
Assumptions
•
The light falls square on (along the normal) a flat, clean boundary between two clear, non-magnetic media.
•
E and B must match across the boundary, so the reflected wave’s amplitude is (n₁ − n₂) ÷ (n₁ + n₂) of the incoming one; the intensity goes with its square.
•
Nothing is absorbed: the reflected and transmitted fractions add to 1, and R is the same from either side.
Step-by-step
WE KNOW
n₁ = 1, n₂ = 1.5
FIND
reflected fraction (R), transmitted fraction (T)
Step 1 · Find reflected fraction (R)
Use R = ((n₁ − n₂) ÷ (n₁ + n₂))²
The
reflected
amplitude
is
n₁
−
n₂
n₁
+
n₂
of
the
incoming
one;
square
it
for
the
intensity.
R
=
(
1 − 1.5
1 + 1.5
)²
R
=
(−
0.5
2.5
)
2
R
=
(−0.2)
2
R
=
0.04
Step 2 · Find transmitted fraction (T)
Use T = 1 − R
Nothing is absorbed, so whatever is not reflected goes through.
T
=
1
−
0.04
T
=
0.96
Check
Put the numbers back in. Both sides should match.
0.04
=
(
1 − 1.5
1 + 1.5
)²
✓
0.96 = 1 − 0.04 ✓
Topic check
See what you know: 5 questions on this topic
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