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Edexcel A Level (IAL) Physics-2.14 Critical Angle- Study Notes- New Syllabus

Edexcel A Level (IAL) Physics -2.14 Critical Angle- Study Notes- New syllabus

Edexcel A Level (IAL) Physics -2.14 Critical Angle- Study Notes -Edexcel A level Physics – per latest Syllabus.

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Critical Angle and Total Internal Reflection \( \sin C = \frac{1}{n} \)

When light travels from a denser medium (higher refractive index) to a less dense medium (lower refractive index), the angle of refraction increases. Beyond a certain incident angle, refraction no longer occurs — the light is totally reflected. This incident angle is the critical angle.

What is the Critical Angle?

The critical angle \( C \) is the angle of incidence in a denser medium for which the angle of refraction is \( 90^\circ \).

Occurs only when:

  • \( n_1 > n_2 \) (light going from denser to rarer medium)
  • Refracted ray travels along the boundary

Formula for the Critical Angle

From Snell’s law:

$ n_1 \sin C = n_2 \sin 90^\circ $

Since \( \sin 90^\circ = 1 \), and if the less dense medium is air \( (n_2 = 1) \):

$ \sin C = \frac{1}{n} $

  • \( C \) = critical angle
  • \( n \) = refractive index of the denser medium

Total Internal Reflection (TIR)

Total internal reflection occurs when:

  • Light travels from higher \( n \) → lower \( n \)
  • Angle of incidence > critical angle

Effects:

  • No refracted ray
  • All light is reflected back into the denser medium
  • Reflection obeys the normal law of reflection

Applications:

  • Optical fibres
  • Endoscopes
  • Prism binoculars
  • Diamond sparkle (high refractive index → small critical angle)

Using \( \sin C = \frac{1}{n} \)

Given the refractive index of a medium, the critical angle can be calculated by:

$ C = \sin^{-1}\left(\frac{1}{n}\right) $

Notes:

  • Applies only when light is leaving the denser medium.
  • If \( n \leq 1 \), no critical angle exists.
  • Higher refractive index → smaller critical angle.

Example (Easy)

Diamond has refractive index \( n = 2.42 \). Find the critical angle at the diamond–air boundary.

▶️ Answer / Explanation

$ \sin C = \frac{1}{n} = \frac{1}{2.42} = 0.413 $

$ C = \sin^{-1}(0.413) \approx 24.4^\circ $

Critical angle ≈ \( 24^\circ \)

Example (Medium)

A transparent plastic has refractive index \( n = 1.50 \). Calculate the critical angle.

▶️ Answer / Explanation

$ \sin C = \frac{1}{1.50} = 0.667 $ $ C = \sin^{-1}(0.667) \approx 41.8^\circ $

Critical angle ≈ \( 42^\circ \)

Example (Hard)

A ray travels from glass (refractive index 1.60) into water (refractive index 1.33). Find the critical angle at this boundary.

▶️ Answer / Explanation

The formula \( \sin C = \frac{1}{n} \) applies only when the less dense medium is air. Here we must use Snell’s law because both media have refractive indices > 1.

Step 1: For critical angle, refracted angle = 90°

$ n_1 \sin C = n_2 \sin 90^\circ $

$ 1.60 \sin C = 1.33 $

Step 2: Solve

$ \sin C = \frac{1.33}{1.60} = 0.831 $ $ C = \sin^{-1}(0.831) \approx 56.3^\circ $

Critical angle ≈ \( 56^\circ \)

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