Physics by Lamhi: Not Your Boring Physics

CLASS 12 · CHAPTER 9 · OPTICS

Ray Optics and Optical Instruments

Increase the angle of incidence inside glass, and refraction bends further and further from the normal, right up until one exact angle. Cross it, and refraction doesn't fade, it switches off completely, like a light switch.

01 · See it

Watch it happen

Below the critical angle, light refracts out into the air as usual, bending away from the normal.

30.0°
Angle of incidence θ₁
48.6°
Angle of refraction θ₂
Refracts
Status
41.81°
Critical angle θc

Drag the slider from 0° upward. The refracted ray in air bends further from the normal as θ₁ grows, exactly as Snell’s law predicts, right up to θc ≈ 41.81°. The moment you cross that one angle, the refracted ray disappears entirely and the status readout flips to Total Internal Reflection: there is no gradual fade, only a hard discontinuity at exactly θc.

02 · Derive it

Where the formula comes from

Snell’s law governs refraction at any boundary between two media of refractive index n1n_1 and n2n_2:

Snell's law
n1sin⁡θ1=n2sin⁡θ2\textcolor{#e08a1e}{n_1\sin\theta_1} = n_2\sin\theta_2

Going from a denser medium (n1n_1, e.g. glass) toward a rarer one (n2n_2, e.g. air), so n1>n2n_1 > n_2, solving for sin⁡θ2\sin\theta_2 gives:

sin⁡θ2=n1n2sin⁡θ1\sin\theta_2 = \dfrac{n_1}{n_2}\sin\theta_1

Since n1/n2>1n_1/n_2 > 1, the right side grows faster than sin⁡θ1\sin\theta_1 itself. The critical angle θc\theta_c is defined as the one angle of incidence at which the refracted ray grazes exactly along the boundary, θ2=90∘\theta_2 = 90^\circ:

At the critical angle
n1sin⁡θc=n2sin⁡90∘=n2n_1\sin\theta_c = n_2\sin 90^\circ = n_2
Critical angle
sin⁡θc=n2n1\textcolor{#e08a1e}{\sin\theta_c = \dfrac{n_2}{n_1}}

For any θ1>θc\theta_1 > \theta_c, the equation sin⁡θ2=(n1/n2)sin⁡θ1\sin\theta_2 = (n_1/n_2)\sin\theta_1 would demand sin⁡θ2>1\sin\theta_2 > 1, which is impossible. No refracted ray can exist, so all the light reflects back into the denser medium: total internal reflection.

03 · Break it

Where the shortcut stops working

Most students picture refraction “weakening” as the angle of incidence grows, until it eventually stops. That picture is wrong, and the simulation above is built to make the wrongness visible: refraction does not weaken, it is simply present, right up to θ1=θc\theta_1 = \theta_c, and then completely absent for every single angle beyond it.

At θ1\theta_1 just below θc\theta_c, a real refracted ray exists, grazing almost flat along the surface at nearly 90°. At θ1=θc\theta_1 = \theta_c exactly, that ray grazes at precisely 90∘90^\circ. One hundredth of a degree further, and Snell’s law has no real solution at all, there is no angle whose sine exceeds 1. The refracted ray does not get fainter or more grazing still, it simply ceases to exist, and 100% of the light reflects back internally instead. This is a genuine discontinuity in the physics, not a smooth transition, and exam questions are written specifically to catch students who assume otherwise.

This one discontinuity is the entire physical basis of optical fibres, where light launched at a steep angle keeps re-striking the core-cladding boundary beyond θc at every bounce, so it is reflected, never refracted out, carrying a signal the length of the fibre with almost no loss. It is also why diamonds sparkle: a diamond’s very high refractive index gives it an unusually small critical angle (≈24.4°), so light entering it keeps exceeding that small θc at its internal facets, bouncing around via total internal reflection several times before it finally escapes.

04 · Master it

Apply it under exam conditions

Q1. Find the critical angle for light travelling from water (n = 1.33) toward air, and the angle of refraction for a ray inside the water incident at 20° (well below the critical angle).

sin⁡θc=11.33=0.7519  ⇒  θc=48.75∘\sin\theta_c = \dfrac{1}{1.33} = 0.7519 \;\Rightarrow\; \theta_c = \textcolor{#e08a1e}{48.75^\circ}sin⁡θ2=1.33sin⁡20∘=1.33(0.342)=0.4549  ⇒  θ2=27.06∘\sin\theta_2 = 1.33\sin 20^\circ = 1.33(0.342) = 0.4549 \;\Rightarrow\; \theta_2 = \textcolor{#e08a1e}{27.06^\circ}

Since 20° is far below θc ≈ 48.75°, a normal refracted ray exists, bent away from the normal as expected.

Q2. A certain glass has refractive index 1.6. Find its critical angle, and determine what happens to a ray inside the glass incident on the glass-air boundary at 25° versus at 45°.

sin⁡θc=11.6=0.625  ⇒  θc=38.68∘\sin\theta_c = \dfrac{1}{1.6} = 0.625 \;\Rightarrow\; \theta_c = \textcolor{#e08a1e}{38.68^\circ}

At 25° (below θc), refraction occurs normally:

sin⁡θ2=1.6sin⁡25∘=1.6(0.4226)=0.6762  ⇒  θ2=42.55∘\sin\theta_2 = 1.6\sin 25^\circ = 1.6(0.4226) = 0.6762 \;\Rightarrow\; \theta_2 = \textcolor{#e08a1e}{42.55^\circ}

At 45° (above θc = 38.68°), sin⁡θ2\sin\theta_2 would have to equal 1.6sin⁡45∘=1.1311.6\sin45^\circ = 1.131, impossible. The ray undergoes total internal reflection instead, with no refracted ray at all.

05 · FAQs

Quick answers

Does refraction just get weaker and weaker as I approach the critical angle, until it fades out?+

No, and this is the single most common misreading of this topic. Right up to and including the critical angle itself, a real refracted ray exists, it's just increasingly grazing, bent almost flat along the surface. The instant the angle of incidence crosses θc, the refracted ray doesn't fade, it vanishes completely and discontinuously, because Snell's law literally has no solution (sinθ₂ would have to exceed 1). It's an on/off switch at one exact angle, not a dimmer.

Does total internal reflection happen at any boundary between two media?+

No, it only happens going from a denser medium (higher n) toward a rarer one (lower n), and only beyond that pair's specific critical angle. Light going from a rarer medium into a denser one always refracts, no matter the angle, since Snell's law n1 sinθ1 = n2 sinθ2 always has a valid solution for sinθ2 when n1 < n2.

Why do diamonds sparkle so much more than glass?+

A diamond's very high refractive index (≈2.42) gives it a very small critical angle (≈24.4°), far smaller than glass's ≈41.8°. Light entering a cut diamond strikes its internal facets at angles that are very likely to exceed this small θc, so it undergoes total internal reflection many times before finally exiting, scattering out as the bright flashes you see, rather than leaking straight through on the first hit.

How does this make optical fibres possible?+

An optical fibre's core has a slightly higher refractive index than its outer cladding. Light launched into the core at a steep enough angle keeps striking the core-cladding boundary beyond the critical angle, so it reflects internally, bounce after bounce, and travels the entire length of the fibre with essentially no loss through the sides, only at the very ends.