Refraction and critical angle: measure from the normal
Before substituting into a refraction equation, draw the boundary, identify the incident medium and mark the normal. Many wrong answers come from using the angle to the surface or reversing the two media. A correct calculator operation cannot repair a wrongly labelled diagram.
These worked examples of refraction and critical angle use a flat boundary between transparent, uniform media. Indices are supplied for the same light colour. Treat air as having refractive index 1.00. Set the calculator to degrees. The sketches are schematic and not to scale; the stated angles, media and ray directions supply the exact data.
Angles and indices: establish the model
The normal is an imaginary line perpendicular to the boundary at the point where the ray meets it. The angle of incidence i is between the incident ray and the normal in the incident medium. The angle of refraction r is between the transmitted ray and the normal in the second medium. If an angle α is measured to the surface, its angle to the normal is 90° − α.
Refractive index n = c/v compares light speed in vacuum with speed in the medium. It has no unit. A higher index means a lower light speed for the stated conditions. “Optically denser” refers here to refractive index, not directly to mass density.
Snell’s law is n1 sin i = n2 sin r, where medium 1 contains the approaching ray and medium 2 contains the transmitted ray. Light entering a higher-index medium bends towards the normal, so r is smaller than i. Entering a lower-index medium, a transmitted ray bends away from the normal.
At normal incidence, i = 0° and r = 0°. The speed can change even though the direction does not. Frequency remains unchanged at a stationary boundary; wavelength changes with speed. This explains why “no bending” does not imply that the two materials have the same refractive index.
Worked example 1: an angle given to the surface
A ray travels from air, n1 = 1.00, into glass, n2 = 1.50. Its incoming angle to the horizontal surface is 30°. Find the angle of refraction.
normal
|
air i = 60° |
ray ↘ |
--------------------O---------------- boundary
glass | ↘ transmitted ray
| r = 35.3°
The given 30° is not i. Therefore i = 90° − 30° = 60°. Snell’s law gives sin r = (1.00/1.50) sin 60° = 0.577350. Taking the inverse sine gives r = 35.264°, or 35.3° to one decimal place.
The transmitted angle is smaller than 60°, consistent with bending towards the normal. If asked for the refracted ray’s angle to the surface, calculate 90° − 35.264° = 54.736°, or 54.7°. These are complementary descriptions of the same ray, not competing answers.
Worked example 2: find an unknown index
Light travels from air into a transparent material. The measured incident and refracted angles, both to the normal, are 45.0° and 28.0°. Determine the material’s refractive index to three significant figures.
Rearrange before evaluating: n2 = n1 sin i/sin r = 1.00 × sin 45.0°/sin 28.0° = 1.50618. Hence n2 = 1.51. There is no degree sign on this result because an index is a ratio, not an angle.
Check the assignment: the ray bends towards the normal, so the second index should be greater than the first. Reversing the sine ratio would give approximately 0.664, contradicting the stated direction of bending in this ordinary transparent-material model.
Do not calculate sin(i/r). The two sine values are calculated separately and then divided. Keep calculator precision until the final result; rounding both sines to one decimal place would introduce an avoidable error.
Critical angle and total internal reflection
A critical angle exists for incidence from a higher-index medium into a lower-index medium. At i = c, the refracted ray lies along the boundary, so r = 90°. Snell’s law becomes sin c = n2/n1, with n1 greater than n2.
Total internal reflection requires both conditions: the ray is in the higher-index medium approaching the lower-index medium, and its incidence angle is greater than the critical angle. At the critical angle itself, use the limiting refracted ray along the surface. Do not replace “greater than” by “greater than or equal to” when classifying that boundary case in a ray diagram.
Worked example 3: glass to air
A ray inside glass of index 1.50 approaches air of index 1.00. The critical angle is c = sin−1(1.00/1.50) = 41.810°, approximately 41.8°.
normal
|
glass c = 41.8° |
incoming ↘ |
---------------------O----------------→ limiting ray
air | r = 90°
At i = 30°, sin r = 1.50 sin 30° = 0.750, so r ≈ 48.6° and refraction occurs. At i = c, the limiting refracted angle is 90°. At i = 50°, the incidence angle exceeds c, so total internal reflection occurs.
normal
|
glass incoming ↘ | ↗ reflected
i=50° | 50°
---------------------O---------------- boundary
air | no transmitted ray
A calculator attempt gives sin r = 1.50 sin 50° ≈ 1.149. Since a real angle cannot have sine greater than one, there is no transmitted ray satisfying the equation. This is a physical clue, not a reason to round the value down to one.
Worked example 4: the second medium matters
Now the same glass meets water of index 1.33. The critical angle becomes c = sin−1(1.33/1.50) = 62.457°, approximately 62.5°. A 50° incident ray that was totally internally reflected at glass–air can now refract into water because 50° is below this new critical angle.
For this glass–water ray, sin r = (1.50/1.33) sin 50° = 0.863960, giving r ≈ 59.8°. It bends away from the normal. Using sin c = 1/n is valid only when the second index is approximated by one; it is not the general critical-angle formula.
Three solved practice tasks
Practice 1: air into water
A ray enters water of index 1.33 from air at 40° to the normal. Find the refracted angle.
Solution: sin r = sin 40°/1.33 = 0.483299, so r ≈ 28.9°. Because water has the higher index, the smaller refracted angle is physically sensible.
Practice 2: test a false reflection claim
A learner says that a ray entering glass from air at 70° must undergo total internal reflection because 70° exceeds 41.8°. Explain the error and find r for glass index 1.50.
Solution: The ray approaches from the lower-index medium, so the direction condition fails. The 41.8° critical angle applies to glass-to-air incidence. Here sin r = sin 70°/1.50 = 0.626462, and r ≈ 38.8°. A transmitted ray exists.
Practice 3: use a surface angle at the critical boundary
Inside a material, a ray reaches an air boundary at 48° to the surface and emerges along the boundary. Find the material’s index.
Solution: The critical angle to the normal is 90° − 48° = 42°. Thus n = 1/sin 42° ≈ 1.49. Using 48° as the critical angle would answer a different geometry.
Frequently asked questions
What is the normal?
It is a perpendicular construction line at the point of incidence. It is not the surface and is not a separate light ray.
Which angle goes into Snell’s law?
Use angles to the normal on the appropriate sides of the boundary. Convert a surface angle before substituting.
Can air-to-glass light undergo total internal reflection?
Not at that ordinary boundary in that direction. Total internal reflection requires incidence from higher to lower refractive index.
Is there reflection below the critical angle?
There can be partial reflection as well as transmission. A diagram focusing on the refracted ray need not show every weaker reflected component.
Does the critical angle belong to one material alone?
No. It depends on the pair of media and the light colour. A quoted material-to-air critical angle assumes air is the second medium.
Why can a calculation return an invalid inverse sine?
Check angle conventions and medium order first. For correctly assigned high-to-low incidence, a sine above one indicates total internal reflection.
Further revision
The existing Physics topical revision collection requires login for resource contents. The underlying relationships are described in OpenStax’s refraction section and total internal reflection section.
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