The Critical Angle Formula, expressed as \ ( \theta_c = \sin^{-1}\left(\frac{n_r}{n_i}\right) \), defines the minimum angle of incidence beyond which total internal reflection occurs at the boundary between two optical media. This formula is a core concept in NCERT Class 12 Physics, Chapter 9 (Ray Optics and Optical Instruments), and it appears regularly in CBSE board exams as well as JEE Main, JEE Advanced, and NEET. In this article, you will find the complete derivation, a ready-to-use formula sheet, three progressive solved examples, CBSE exam tips for 2025-26, common mistakes, and JEE/NEET application patterns.
Key Critical Angle Formulas at a Glance
Quick reference for the most important formulas on critical angle and total internal reflection.
- Critical angle: \( \theta_c = \sin^{-1}\left(\dfrac{n_r}{n_i}\right) \)
- Snell's Law at critical angle: \( n_i \sin\theta_c = n_r \sin 90^\circ \)
- For glass-air interface: \( \theta_c = \sin^{-1}\left(\dfrac{1}{n}\right) \)
- Refractive index relation: \( n = \dfrac{1}{\sin\theta_c} \)
- Speed of light relation: \( \sin\theta_c = \dfrac{v_i}{v_r} \)
- Wavelength relation: \( \sin\theta_c = \dfrac{\lambda_i}{\lambda_r} \)
What is the Critical Angle Formula?
The Critical Angle Formula gives the specific angle of incidence at which a ray of light, travelling from a denser medium to a rarer medium, refracts at exactly 90° along the interface. When the angle of incidence exceeds this critical value, no refraction occurs. Instead, the light bounces entirely back into the denser medium. This phenomenon is called total internal reflection.
According to NCERT Class 12 Physics, Chapter 9, the critical angle depends only on the refractive indices of the two media involved. A higher refractive index of the denser medium means a smaller critical angle. For example, diamond has a very high refractive index of about 2.42, giving it a critical angle of only 24.4°. This is why diamonds sparkle so brilliantly.
The concept is also foundational for understanding optical fibres, periscopes, binoculars, and mirage formation. CBSE Class 12 students must know this formula for both the theory and numerical sections of their board exam. JEE and NEET aspirants encounter it in optics problems involving prisms and fibre optics.
Critical Angle Formula — Expression and Variables
The Critical Angle Formula is derived directly from Snell's Law. When the refracted ray grazes the interface, the angle of refraction becomes 90°. Substituting this condition into Snell's Law yields:
\[ \theta_c = \sin^{-1}\left(\frac{n_r}{n_i}\right) \]
Here, \( \theta_c \) is the critical angle, \( n_r \) is the refractive index of the rarer (second) medium, and \( n_i \) is the refractive index of the denser (incident) medium. The formula requires \( n_i > n_r \) for a critical angle to exist.
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( \theta_c \) | Critical angle | Degrees (°) or Radians (rad) |
| \( n_i \) | Refractive index of the incident (denser) medium | Dimensionless |
| \( n_r \) | Refractive index of the refracted (rarer) medium | Dimensionless |
| \( \sin^{-1} \) | Inverse sine (arcsine) function | — |
Derivation of the Critical Angle Formula
The derivation starts with Snell's Law at the interface of two media:
\[ n_i \sin\theta_i = n_r \sin\theta_r \]
Step 1: At the critical angle, the refracted ray travels along the interface. Therefore, \( \theta_r = 90^\circ \).
Step 2: Substitute \( \theta_r = 90^\circ \) into Snell's Law:
\[ n_i \sin\theta_c = n_r \sin 90^\circ \]
Step 3: Since \( \sin 90^\circ = 1 \):
\[ n_i \sin\theta_c = n_r \]
Step 4: Rearranging for \( \theta_c \):
\[ \theta_c = \sin^{-1}\left(\frac{n_r}{n_i}\right) \]
This is the standard Critical Angle Formula used in NCERT Class 12 Physics and all competitive exams.
Complete Ray Optics Formula Sheet
The table below covers all key formulas from the Ray Optics chapter in NCERT Class 12 Physics. Use it as a quick revision reference before your CBSE board exam or JEE/NEET test.
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Critical Angle Formula | \( \theta_c = \sin^{-1}\left(\dfrac{n_r}{n_i}\right) \) | n_r = rarer medium index, n_i = denser medium index | Degrees or rad | Class 12, Ch 9 |
| Snell's Law | \( n_1 \sin\theta_1 = n_2 \sin\theta_2 \) | n = refractive index, θ = angle with normal | Dimensionless | Class 12, Ch 9 |
| Refractive Index (speed) | \( n = \dfrac{c}{v} \) | c = speed of light in vacuum, v = speed in medium | Dimensionless | Class 12, Ch 9 |
| Refractive Index from Critical Angle | \( n = \dfrac{1}{\sin\theta_c} \) | n = refractive index of denser medium (rarer = air) | Dimensionless | Class 12, Ch 9 |
| Mirror Formula | \( \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} \) | f = focal length, v = image distance, u = object distance | m | Class 12, Ch 9 |
| Lens Maker's Formula | \( \dfrac{1}{f} = (n-1)\left(\dfrac{1}{R_1} – \dfrac{1}{R_2}\right) \) | n = refractive index, R_1, R_2 = radii of curvature | m | Class 12, Ch 9 |
| Thin Lens Formula | \( \dfrac{1}{f} = \dfrac{1}{v} – \dfrac{1}{u} \) | f = focal length, v = image distance, u = object distance | m | Class 12, Ch 9 |
| Magnification (Lens) | \( m = \dfrac{v}{u} \) | v = image distance, u = object distance | Dimensionless | Class 12, Ch 9 |
| Angle of Deviation (Prism) | \( \delta = (n-1)A \) | n = refractive index, A = prism angle | Degrees | Class 12, Ch 9 |
| Numerical Aperture (Optical Fibre) | \( NA = \sqrt{n_1^2 – n_2^2} \) | n_1 = core index, n_2 = cladding index | Dimensionless | Class 12, Ch 9 |
Critical Angle Formula — Solved Examples
Example 1 (Class 9-10 Level — Direct Application)
Problem: The refractive index of glass is 1.5. Calculate the critical angle for a glass-air interface.
Given: \( n_i = 1.5 \) (glass), \( n_r = 1.0 \) (air)
Step 1: Write the Critical Angle Formula:
\( \theta_c = \sin^{-1}\left(\dfrac{n_r}{n_i}\right) \)
Step 2: Substitute the known values:
\( \theta_c = \sin^{-1}\left(\dfrac{1.0}{1.5}\right) = \sin^{-1}(0.6667) \)
Step 3: Evaluate using inverse sine:
\( \theta_c \approx 41.8^\circ \)
Answer
The critical angle for the glass-air interface is approximately 41.8°. Any ray striking the glass-air boundary at an angle greater than 41.8° undergoes total internal reflection.
Example 2 (Class 11-12 Level — Multi-Step)
Problem: A ray of light travels from a medium of refractive index 1.67 into air. Find the critical angle. Also determine the speed of light in the denser medium. (Speed of light in vacuum \( c = 3 \times 10^8 \) m/s)
Given: \( n_i = 1.67 \), \( n_r = 1.0 \) (air), \( c = 3 \times 10^8 \) m/s
Step 1: Apply the Critical Angle Formula:
\( \theta_c = \sin^{-1}\left(\dfrac{1}{1.67}\right) = \sin^{-1}(0.5988) \)
Step 2: Evaluate the inverse sine:
\( \theta_c \approx 36.8^\circ \approx 0.642 \) rad
Step 3: Find the speed of light in the medium using \( n = c/v \):
\( v = \dfrac{c}{n_i} = \dfrac{3 \times 10^8}{1.67} \approx 1.796 \times 10^8 \) m/s
Answer
Critical angle \( \theta_c \approx 36.8^\circ \). Speed of light in the medium \( \approx 1.80 \times 10^8 \) m/s.
Example 3 (JEE/NEET Level — Concept Application)
Problem: An optical fibre has a core refractive index of 1.62 and a cladding refractive index of 1.52. (a) Find the critical angle at the core-cladding interface. (b) A ray enters the fibre at an angle of 30° to the axis. Does it undergo total internal reflection at the core-cladding boundary?
Given: \( n_{core} = 1.62 \), \( n_{clad} = 1.52 \), angle of ray with fibre axis = 30°
Step 1: Calculate the critical angle using the Critical Angle Formula:
\( \theta_c = \sin^{-1}\left(\dfrac{n_{clad}}{n_{core}}\right) = \sin^{-1}\left(\dfrac{1.52}{1.62}\right) = \sin^{-1}(0.9383) \)
Step 2: Evaluate:
\( \theta_c \approx 69.7^\circ \)
Step 3: Find the angle of incidence at the core-cladding boundary. The ray makes 30° with the fibre axis, so it makes \( 90^\circ – 30^\circ = 60^\circ \) with the normal to the boundary.
Step 4: Compare the angle of incidence (60°) with the critical angle (69.7°).
Since \( 60^\circ < 69.7^\circ \), the angle of incidence is less than the critical angle.
Answer
(a) Critical angle \( \approx 69.7^\circ \). (b) The ray does not undergo total internal reflection at this boundary because its angle of incidence (60°) is less than the critical angle. The ray will refract out of the core. For total internal reflection, the ray must hit the boundary at an angle greater than 69.7°.
CBSE Exam Tips 2025-26
- Always state Snell's Law first. In your CBSE 2025-26 answer, write Snell's Law before deriving the critical angle formula. Examiners award marks for each step of the derivation.
- Remember the condition. The Critical Angle Formula applies only when light travels from a denser medium to a rarer medium (\( n_i > n_r \)). State this condition explicitly in your answer.
- Learn key values by heart. For glass (n = 1.5), \( \theta_c \approx 41.8^\circ \). For diamond (n = 2.42), \( \theta_c \approx 24.4^\circ \). These appear directly in CBSE numericals.
- Link to applications. CBSE 3-mark questions often ask you to name two applications of total internal reflection. We recommend memorising: optical fibres, diamond brilliance, mirage, periscope, and binoculars.
- Unit awareness. The critical angle is a pure angle. Express it in degrees unless the question asks for radians. Refractive index has no unit — do not write “m/s” or any unit next to it.
- Check your calculator mode. In board exams, always confirm your calculator is in degree mode before computing \( \sin^{-1} \). A common error is getting the answer in radians by mistake.
Common Mistakes to Avoid
- Inverting the ratio. Many students write \( \theta_c = \sin^{-1}(n_i / n_r) \) instead of \( \sin^{-1}(n_r / n_i) \). Remember: the rarer medium index goes on top. The ratio must always be less than 1 for a valid critical angle.
- Applying the formula in the wrong direction. The Critical Angle Formula is only valid for light going from a denser medium to a rarer medium. If light travels from air to glass, there is no critical angle.
- Confusing angle of incidence with angle from the surface. The angle in the formula is always measured from the normal to the surface, not from the surface itself. In fibre optics problems, convert carefully.
- Ignoring the condition \( n_i > n_r \). If \( n_r / n_i > 1 \), the arcsine is undefined. Always verify that the incident medium is denser before applying the formula.
- Mixing up refractive index with speed. A higher refractive index means a lower speed of light in that medium. Do not confuse the two when comparing media.
JEE/NEET Application of Critical Angle Formula
In our experience, JEE aspirants encounter the Critical Angle Formula in at least one question per year, typically embedded in a prism or optical fibre problem. NEET questions tend to be more direct, testing the formula and its applications. Here are the three most common patterns:
Pattern 1: Finding Critical Angle from Refractive Index
JEE Main frequently gives the refractive index of a medium and asks for the critical angle. The calculation is direct. However, JEE Advanced may give the refractive index as a function of wavelength (\( n = A + B/\lambda^2 \)) and ask how the critical angle changes with colour. Since violet light has a higher refractive index than red light, it has a smaller critical angle. Our experts suggest practising at least five such wavelength-dependent problems before your JEE exam.
Pattern 2: Prism Problems Involving Total Internal Reflection
A common JEE problem involves a prism where one face causes total internal reflection. You must first find the angle of incidence at that face using geometry, then compare it to the critical angle. If the angle of incidence exceeds the critical angle, the ray is totally reflected. The combined use of the Critical Angle Formula and prism geometry is a high-weightage JEE topic.
Pattern 3: Optical Fibre and Numerical Aperture
NEET and JEE both ask about optical fibres. The acceptance angle of a fibre is directly related to the critical angle at the core-cladding interface. The numerical aperture \( NA = \sqrt{n_1^2 – n_2^2} \) can also be expressed as \( NA = n_0 \sin\theta_{accept} \), where \( \theta_{accept} \) is the maximum acceptance angle. Mastering the Critical Angle Formula is the first step to solving these problems confidently.
FAQs on Critical Angle Formula
Explore more optics and physics formulas on our Physics Formulas hub. For related topics, study the Angular Velocity Formula for rotational optics problems, and the Fluid Mechanics Formula page for wave-medium interaction concepts. You can also verify the NCERT syllabus for Class 12 Ray Optics at the official NCERT website. For more formula practice, see our Normal Force Formula and Instantaneous Speed Formula articles.