The Brewsters Law Formula states that the tangent of the polarising angle equals the refractive index of the medium, expressed as \( \mu = an( heta_B) \), and it is a fundamental concept covered in Class 12 Physics under the chapter on Wave Optics. This formula is essential for CBSE board exams and appears regularly in JEE Main and NEET papers. Understanding Brewster's Law helps students solve problems involving polarisation of light, reflected and refracted rays, and optical properties of materials. This article covers the complete formula, derivation, variables, a full formula sheet, three solved examples at progressive difficulty levels, CBSE exam tips for 2025-26, common mistakes, and JEE/NEET applications.
Key Brewsters Law Formulas at a Glance
Quick reference for the most important Brewster's Law formulas and related polarisation expressions.
- Brewster's Law: \( \mu = an( heta_B) \)
- Snell's Law at Brewster's angle: \( n_1 \sin( heta_B) = n_2 \sin( heta_r) \)
- Angle condition: \( heta_B + heta_r = 90° \)
- Brewster's angle for air-glass interface: \( heta_B = an^{-1}(1.5) \approx 56.3° \)
- Degree of polarisation: \( P = rac{I_{max} – I_{min}}{I_{max} + I_{min}} \)
- Malus's Law: \( I = I_0 \cos^2( heta) \)
- Refractive index relation: \( \mu = rac{c}{v} \)
What is Brewsters Law Formula?
The Brewsters Law Formula describes the relationship between the polarising angle and the refractive index of a medium. When unpolarised light strikes a surface at a specific angle called the Brewster angle (or polarising angle), the reflected light becomes completely plane-polarised. At this angle, the reflected and refracted rays are perpendicular to each other.
This law was discovered by Scottish physicist Sir David Brewster in 1815. It is covered in Class 12 Physics, Chapter 10 — Wave Optics, as per the NCERT curriculum. The law connects two fundamental optical quantities: the angle of incidence and the refractive index of the denser medium.
The significance of this law lies in its practical applications. Polaroid sunglasses, LCD screens, photography filters, and laser technology all rely on Brewster's principle. For CBSE students, this formula is a high-scoring topic. For JEE and NEET aspirants, it combines with Snell's Law and Malus's Law to form complex multi-step problems.
The key insight is simple: at Brewster's angle, the angle between the reflected ray and the refracted ray is exactly 90 degrees. This geometric condition leads directly to the elegant formula \( \mu = an( heta_B) \).
Brewsters Law Formula — Expression and Variables
The mathematical statement of Brewster's Law is:
\[ \mu = \tan(\theta_B) \]
where \( \mu \) is the refractive index of the medium and \( \theta_B \) is the Brewster angle (polarising angle).
An equivalent form using Snell's Law is:
\[ \tan(\theta_B) = \frac{n_2}{n_1} \]
This applies when light travels from medium 1 (refractive index \( n_1 \)) into medium 2 (refractive index \( n_2 \)).
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( \mu \) or \( n \) | Refractive index of the medium | Dimensionless |
| \( \theta_B \) | Brewster's angle (polarising angle) | Degrees or Radians |
| \( n_1 \) | Refractive index of incident medium | Dimensionless |
| \( n_2 \) | Refractive index of refracting medium | Dimensionless |
| \( \theta_r \) | Angle of refraction at Brewster's angle | Degrees or Radians |
| \( c \) | Speed of light in vacuum | m/s |
| \( v \) | Speed of light in the medium | m/s |
Derivation of Brewsters Law Formula
The derivation starts from the geometric condition at Brewster's angle. At this angle, the reflected and refracted rays are perpendicular:
Step 1: State the perpendicularity condition: \( \theta_B + \theta_r = 90° \), so \( \theta_r = 90° – \theta_B \).
Step 2: Apply Snell's Law: \( n_1 \sin(\theta_B) = n_2 \sin(\theta_r) \).
Step 3: Substitute \( \theta_r = 90° – \theta_B \): \( n_1 \sin(\theta_B) = n_2 \sin(90° – \theta_B) = n_2 \cos(\theta_B) \).
Step 4: Rearrange: \( \frac{n_2}{n_1} = \frac{\sin(\theta_B)}{\cos(\theta_B)} = \tan(\theta_B) \).
Step 5: For light incident from air (\( n_1 = 1 \)): \( \mu = \tan(\theta_B) \).
This completes the derivation. The result is elegant because it directly links a measurable angle to the optical property of the medium.
Complete Wave Optics Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Brewster's Law | \( \mu = \tan(\theta_B) \) | \( \mu \)=refractive index, \( \theta_B \)=Brewster's angle | Dimensionless | Class 12, Ch 10 |
| Snell's Law | \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \) | \( n \)=refractive index, \( \theta \)=angle | Dimensionless | Class 12, Ch 9 |
| Malus's Law | \( I = I_0 \cos^2(\theta) \) | \( I \)=transmitted intensity, \( I_0 \)=initial intensity, \( \theta \)=angle between polariser axes | W/m² | Class 12, Ch 10 |
| Degree of Polarisation | \( P = \frac{I_{max} – I_{min}}{I_{max} + I_{min}} \) | \( I_{max} \)=maximum intensity, \( I_{min} \)=minimum intensity | Dimensionless | Class 12, Ch 10 |
| Refractive Index | \( \mu = \frac{c}{v} \) | \( c \)=speed of light in vacuum, \( v \)=speed in medium | Dimensionless | Class 12, Ch 9 |
| Young's Double Slit (fringe width) | \( \beta = \frac{\lambda D}{d} \) | \( \lambda \)=wavelength, \( D \)=screen distance, \( d \)=slit separation | m | Class 12, Ch 10 |
| Condition for Constructive Interference | \( \Delta x = n\lambda \) | \( \Delta x \)=path difference, \( n \)=integer, \( \lambda \)=wavelength | m | Class 12, Ch 10 |
| Condition for Destructive Interference | \( \Delta x = (2n-1)\frac{\lambda}{2} \) | \( \Delta x \)=path difference, \( n \)=integer | m | Class 12, Ch 10 |
| Resolving Power of Telescope | \( \theta_{min} = \frac{1.22\lambda}{D} \) | \( \lambda \)=wavelength, \( D \)=aperture diameter | Radians | Class 12, Ch 10 |
| Brewster's Angle (inverse form) | \( \theta_B = \tan^{-1}(\mu) \) | \( \mu \)=refractive index of medium | Degrees | Class 12, Ch 10 |
Brewsters Law Formula — Solved Examples
Example 1 (Class 11-12 Level — Direct Application)
Problem: Light is incident on a glass surface from air. The refractive index of glass is 1.732. Find the Brewster's angle for this glass surface.
Given: Refractive index of glass, \( \mu = 1.732 \); incident medium is air (\( n_1 = 1 \)).
Step 1: Write Brewster's Law: \( \mu = \tan(\theta_B) \)
Step 2: Substitute the value: \( \tan(\theta_B) = 1.732 \)
Step 3: Take the inverse tangent: \( \theta_B = \tan^{-1}(1.732) \)
Step 4: Calculate: \( \theta_B = 60° \)
Answer
The Brewster's angle for this glass surface is 60°. At this angle, the reflected light will be completely plane-polarised.
Example 2 (Class 12 / CBSE Board Level — Multi-step)
Problem: The Brewster's angle for a glass-air interface is found to be 33.7°. A ray of light is incident from glass onto air at this polarising angle. Find: (a) the refractive index of glass, and (b) the angle of refraction in air.
Given: Brewster's angle from glass to air, \( \theta_B = 33.7° \).
Step 1: Apply Brewster's Law for glass-to-air interface. Here, \( \tan(\theta_B) = \frac{n_{air}}{n_{glass}} \).
Step 2: Calculate: \( \tan(33.7°) \approx 0.667 \). So \( \frac{1}{\mu_{glass}} = 0.667 \).
Step 3: Find refractive index: \( \mu_{glass} = \frac{1}{0.667} \approx 1.5 \).
Step 4: Use the perpendicularity condition to find the angle of refraction: \( \theta_r = 90° – \theta_B = 90° – 33.7° = 56.3° \).
Step 5: Verify using Snell's Law: \( 1.5 \times \sin(33.7°) = 1 \times \sin(56.3°) \). Left side: \( 1.5 \times 0.555 = 0.832 \). Right side: \( \sin(56.3°) \approx 0.832 \). Verified.
Answer
(a) Refractive index of glass = 1.5. (b) Angle of refraction in air = 56.3°. Note that the reflected and refracted rays are perpendicular, confirming Brewster's condition.
Example 3 (JEE/NEET Level — Concept Application)
Problem: Unpolarised light of intensity \( I_0 \) is incident on a glass surface at Brewster's angle. The reflected light passes through a polaroid analyser. The transmission axis of the analyser makes an angle of 30° with the vibration direction of the reflected polarised light. Find the intensity of light transmitted through the analyser. Also, what is the Brewster's angle if the glass has refractive index 1.5?
Given: Incident intensity = \( I_0 \); analyser angle \( \theta = 30° \); \( \mu = 1.5 \).
Step 1: At Brewster's angle, the reflected light is completely plane-polarised. The intensity of completely polarised reflected light from an air-glass interface is approximately \( \frac{I_0}{2} \) (since only the component parallel to the surface is reflected and polarised; for simplicity in this problem, treat reflected polarised intensity as \( I_r \)).
Step 2: For this problem, let the intensity of the completely polarised reflected beam be \( I_r = \frac{I_0}{2} \) (standard approximation used in JEE problems).
Step 3: Apply Malus's Law to find transmitted intensity through the analyser:
\[ I = I_r \cos^2(\theta) = \frac{I_0}{2} \cos^2(30°) \]
Step 4: Calculate: \( \cos(30°) = \frac{\sqrt{3}}{2} \), so \( \cos^2(30°) = \frac{3}{4} \).
Step 5: Therefore: \( I = \frac{I_0}{2} \times \frac{3}{4} = \frac{3I_0}{8} \).
Step 6: Find Brewster's angle for \( \mu = 1.5 \): \( \theta_B = \tan^{-1}(1.5) \approx 56.3° \).
Answer
Intensity transmitted through the analyser = \( \frac{3I_0}{8} \). The Brewster's angle for glass with \( \mu = 1.5 \) is approximately 56.3°. This type of combined Brewster's Law and Malus's Law problem is very common in JEE Main.
CBSE Exam Tips 2025-26
- Memorise the core formula: Write \( \mu = \tan(\theta_B) \) at the top of your answer. CBSE awards one mark simply for stating the correct formula before solving.
- Always state the perpendicularity condition: In 3-mark and 5-mark questions, mention that at Brewster's angle, \( \theta_B + \theta_r = 90° \). This earns a separate step mark.
- Draw a labelled diagram: CBSE 2025-26 marking schemes reward diagrams showing incident ray, reflected ray, refracted ray, normal, and the 90° angle between reflected and refracted rays.
- Link to Snell's Law: Derivation questions require you to start from Snell's Law. We recommend practising the full derivation at least five times before the board exam.
- Know the inverse formula: If a question gives \( \mu \) and asks for \( \theta_B \), use \( \theta_B = \tan^{-1}(\mu) \). Many students forget this rearrangement under exam pressure.
- Check units: The refractive index is dimensionless. The Brewster angle is in degrees. Always state these in your answer to avoid unit-related mark deductions.
Common Mistakes to Avoid
- Mistake 1 — Confusing Brewster's angle with critical angle: The critical angle applies to total internal reflection. Brewster's angle applies to polarisation of reflected light. These are completely different phenomena. Always identify which angle the question is asking about.
- Mistake 2 — Forgetting the 90° condition: Many students write \( \mu = \tan(\theta_B) \) but forget that this comes from \( \theta_B + \theta_r = 90° \). In derivation questions, omitting this step loses marks.
- Mistake 3 — Applying the formula in the wrong direction: When light travels from glass to air, \( \tan(\theta_B) = \frac{n_{air}}{n_{glass}} = \frac{1}{\mu} \), not \( \mu \). Always check the direction of travel before applying the formula.
- Mistake 4 — Mixing up Malus's Law and Brewster's Law: Brewster's Law gives the angle for complete polarisation on reflection. Malus's Law gives the transmitted intensity through an analyser. These are two separate laws used in combination for harder problems.
- Mistake 5 — Using degrees in trigonometric calculators without checking mode: Always ensure your calculator is in degree mode when computing \( \tan^{-1}(\mu) \). A calculator in radian mode gives a completely wrong answer for Brewster's angle.
JEE/NEET Application of Brewsters Law Formula
In our experience, JEE aspirants encounter Brewster's Law in 1-2 questions per year in JEE Main, often combined with Malus's Law or Snell's Law. NEET includes it as a single-concept MCQ from the Wave Optics chapter. Understanding the three key application patterns below will cover most exam scenarios.
Application Pattern 1: Direct Calculation of Brewster's Angle
The most common JEE/NEET question gives the refractive index and asks for the polarising angle. Apply \( \theta_B = \tan^{-1}(\mu) \) directly. Common values to memorise: \( \mu = \sqrt{3} \Rightarrow \theta_B = 60° \); \( \mu = 1 \Rightarrow \theta_B = 45° \); \( \mu = 1.5 \Rightarrow \theta_B \approx 56.3° \).
Application Pattern 2: Combined Brewster's and Malus's Law
JEE Main frequently combines both laws. The setup is: unpolarised light hits a surface at Brewster's angle, the reflected polarised light then passes through a polaroid at angle \( \theta \). The final intensity is \( I = \frac{I_0}{2} \cos^2(\theta) \). The factor of \( \frac{1}{2} \) comes from the fact that only one polarisation component is reflected at Brewster's angle.
Application Pattern 3: Identifying Polarised Light in Multi-Interface Problems
In JEE Advanced, light may pass through multiple surfaces. At each interface, students must check whether the angle of incidence equals the Brewster angle for that interface. If it does, the reflected beam is completely polarised and the refracted beam is partially polarised. Our experts suggest drawing a ray diagram for every multi-interface problem to track the polarisation state at each step.
For NEET, the emphasis is on conceptual understanding: knowing that polarisation is possible only for transverse waves (not longitudinal waves like sound), and that Brewster's Law is evidence for the transverse nature of light. This conceptual point has appeared in NEET multiple times as a one-liner MCQ.
FAQs on Brewsters Law Formula
For more related Physics formulas, explore our comprehensive guides on the Pressure Formula, the Power Formula, and the Average Force Formula. For a complete collection of Physics formulas aligned with the NCERT Class 12 syllabus, visit our Physics Formulas hub. You can also refer to the official NCERT Wave Optics chapter on the NCERT official website for the source textbook content.