Total internal reflection — critical angle and applications in optical fibre

medium CBSE JEE-MAIN NEET CBSE 2024 3 min read

Question

What is total internal reflection? Derive the expression for critical angle. Explain how total internal reflection is used in optical fibres.

(CBSE 2024, 5-mark question)


Solution — Step by Step

When light travels from a denser medium to a rarer medium (e.g., glass to air), and the angle of incidence exceeds a certain value called the critical angle, all light is reflected back into the denser medium. No refraction occurs — this is total internal reflection.

Two conditions for TIR: (1) Light must go from denser to rarer medium. (2) Angle of incidence must exceed the critical angle θc\theta_c.

At the critical angle, the refracted ray grazes along the interface (angle of refraction = 90°). Applying Snell’s law:

n1sinθc=n2sin90°n_1 \sin\theta_c = n_2 \sin 90° n1sinθc=n2n_1 \sin\theta_c = n_2 sinθc=n2n1\sin\theta_c = \frac{n_2}{n_1}

For glass (n1=1.5n_1 = 1.5) to air (n2=1n_2 = 1):

θc=sin1(11.5)=sin1(0.667)42°\theta_c = \sin^{-1}\left(\frac{1}{1.5}\right) = \sin^{-1}(0.667) \approx \mathbf{42°}

An optical fibre consists of a core (higher refractive index) surrounded by cladding (lower refractive index). Light enters the core and hits the core-cladding boundary at an angle greater than the critical angle.

At each reflection point, TIR occurs — no light escapes through the cladding. The light bounces along the fibre, following even curved paths. This allows data transmission over kilometres with minimal signal loss.

  • Core: n11.62n_1 \approx 1.62 (higher)
  • Cladding: n21.52n_2 \approx 1.52 (lower)
  • Acceptance angle: The maximum angle at which light can enter the fibre and still undergo TIR inside

The small difference in refractive indices is carefully engineered — too large and the fibre becomes multimodal (signal dispersion); too small and TIR becomes unreliable.


Why This Works

TIR is a consequence of Snell’s law at its limiting case. When the refracted angle would need to exceed 90° (which is physically impossible), the light has no choice but to reflect entirely. Unlike normal reflection from a mirror (which loses some light), TIR reflects 100% of the incident light — making it extremely efficient.

This is why optical fibres can carry signals across oceans with less loss than copper wires carrying electrical signals across a city.


Alternative Method — Using Relative Refractive Index

If μ\mu is the refractive index of the denser medium relative to the rarer:

sinθc=1μ\sin\theta_c = \frac{1}{\mu}

Other applications of TIR frequently asked in CBSE/NEET: (1) Mirage formation in deserts (layers of air with varying density), (2) Brilliance of diamonds (θc24°\theta_c \approx 24° due to high n=2.42n = 2.42), (3) Totally reflecting prisms used in periscopes and binoculars (replacing mirrors for 100% reflection).


Common Mistake

Students often forget the first condition: light must travel from denser to rarer medium. If light goes from air to glass, TIR is impossible regardless of the angle. In NEET 2023, a question tested exactly this — asking whether TIR can occur when light enters water from air. The answer is no.

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