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Optics
Brewster's Law and Polarization by Reflection
Brewster's law states that when unpolarized light is incident on a transparent dielectric surface at a specific angle called the polarizing angle, the reflected light is completely plane-polarized. At this angle of incidence, the reflected and refracted rays are mutually perpendicular to each other.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Definition of polarizing angle (Brewster angle), statement of Brewster's law, relation between refractive index and tangent of Brewster angle (mu = tan i_p), and the perpendicularity condition between reflected and refracted rays.
PGT extension
Vector formulation of reflection and refraction, phase changes upon reflection at dielectric boundaries, derivation from electromagnetic boundary conditions (Fresnel equations), and behavior of multi-plate glass stacks for polarization.
Prerequisites
Wave nature of light and transverse nature of electromagnetic waves
Snell's law of refraction
Concept of plane-polarized and unpolarized light
Polarization by Reflection
When an unpolarized light beam strikes a smooth interface separating two transparent dielectric media, both reflection and refraction occur.
The reflected beam is partially polarized for all angles of incidence except one specific angle known as the polarizing angle or Brewster's angle (i_p).
At the polarizing angle, the electric field vector oscillating parallel to the plane of incidence is completely transmitted, leaving the reflected ray entirely polarized with vibrations perpendicular to the plane of incidence.
Brewster's Law Formulation
Sir David Brewster discovered experimentally in 1812 that the tangent of the polarizing angle of incidence equals the refractive index of the denser medium with respect to the rarer medium.
Mathematically expressed as tan($i_p) = \mu_21$, where mu_21 is the refractive index of medium 2 with respect to medium 1.
When light travels from air or vacuum (mu = 1) into a medium of refractive index mu, the relation simplifies to tan($i_p) = \mu$.
Geometric Condition at Brewster Angle
At the polarizing angle of incidence i_p, the angle between the reflected ray and the refracted ray is exactly 90 degrees.
Using Snell's law (sin i_p / sin r = mu) and Brewster's law (m$u = \sin i_p / \cos i_p)$, it follows that sin i_p / sin $r = \sin i_p / \cos i_p$.
This implies sin $r = \cos i_p = \sin(90 - i_p)$, leading to $r = 90 - i_p$, or i_p + r = 90 degrees.
Degree of Polarization and Intensity
The reflected intensity is a fraction of the incident unpolarized intensity, and at Brewster's angle, it consists solely of light polarized perpendicular to the plane of incidence.
Brewster's windows are used in lasers to minimize reflection losses for the polarization parallel to the plane of incidence.
Polarization by reflection is independent of the wavelength of light, except for minor variations in refractive index due to dispersion.
Derivations
Derivation of Brewster's Law from Snell's Law
Conditions: Light incident from a medium of refractive index mu_1 to a medium of refractive index mu_2 at polarizing angle i_p.
Apply Snell's law at the interface: mu_1 sin $i_p = \mu_2 \sin r$, where i_p is the polarizing angle and r is the angle of refraction.
State the geometric condition at Brewster's angle: the reflected ray and refracted ray are perpendicular, meaning i_p + r = 90 degrees, or $r = 90 - i_p$.
Substitute r into Snell's law: mu_1 sin $i_p = \mu_2 \sin(90 - i_p) = \mu_2 \cos i_p$.
Rearrange the terms to find the ratio of sine and cosine: sin i_p / cos $i_p = \mu_2 / \mu_1$.
Express as tangent: tan $i_p = \mu_2 / \mu_1 = \mu_21$.
Result: tan $i_p = \mu_21$
Reflected and Refracted Ray Perpendicularity Derivation
Conditions: Plane interface between two dielectric media satisfying Brewster's law.
Let i_p be the angle of incidence equal to Brewster's angle, and r be the angle of refraction.
From Brewster's law, tan $i_p = \mu_2 / \mu_1 = \sin i_p / \cos i_p$.
Equating the two expressions gives sin i_p / cos $i_p = \sin i_p / \sin r$.
Canceling sin i_p yields cos $i_p = \sin r = \cos(90 - r)$, which means $i_p = 90 - r or i_p + r = 90 degrees$.
Since the angle of reflection equals the angle of incidence (i_p), the angle between the reflected ray (at angle i_p to the normal) and the refracted ray (at angle r to the normal) is 180 - (i_p + r) = 180 - 90 = 90 degrees.
Result: Angle between reflected and refracted rays = 90 degrees
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Brewster's Law
$$\tan(i_p) = \frac{\mu_2}{\mu_1}$$
$i_p = polarizing angle$, m$u_1 = refractive index of first medium$, m$u_2 = refractive index of second medium$ Dimensionless
Valid for dielectric, non-magnetic, isotropic media
Brewster's Law for Air-Medium Interface
$$\tan(i_p) = \mu$$
$i_p = polarizing angle$, mu = refractive index of the medium relative to air Dimensionless
Incident medium is air (mu_1 approx 1)
Perpendicularity Condition
$$i_p + r = 90^\circ$$
$i_p = polarizing angle$, r = angle of refraction Degrees or radians
At Brewster's angle
Snell's Law at Brewster Angle
$$\mu = \frac{\sin(i_p)}{\sin(r)} = \tan(i_p)$$
mu = relative refractive index, $i_p = polarizing angle$, r = angle of refraction Dimensionless
Standard refraction at interface
Refractive Index from Polarizing Angle Sine Form
$$\mu = \frac{\sin(i_p)}{\cos(i_p)}$$
mu = refractive index, $i_p = Brewster angle$ Dimensionless
Transition from vacuum/air
Polarizing Angle for Given Refractive Index
$$i_p = \arctan(\mu)$$
$i_p = polarizing angle$, mu = refractive index Radians or degrees
Inverse trigonometric form
Refracted Angle at Brewster Incidence
$$r = 90^\circ - \arctan(\mu)$$
r = angle of refraction, mu = refractive index Degrees
Valid when incident from rarer to denser medium
Brewster Condition in Terms of Phase Velocity
$$\tan(i_p) = \frac{v_1}{v_2}$$
$v_1 = phase velocity in medium 1$, $v_2 = phase velocity in medium 2$ m/s
Derived from m$u = c/v$
Graphs and experiments
Intensity vs Angle of Incidence Graph
Draw: A plot showing reflected light intensity (I_p for parallel polarization and I_s for perpendicular polarization) against the angle of incidence i. I_p drops to zero at $i = i_p$.
Interpret: At Brewster's angle i_p, the component of light polarized parallel to the plane of incidence has zero reflection coefficient. Only the perpendicular component is reflected.
Exam use: Direct identification of Brewster's angle from polarization curves in multi-choice questions.
Reflection and Refraction Setup at Brewster Angle
Draw: A ray diagram showing an unpolarized incident ray hitting a glass slab at angle i_p, a completely polarized reflected ray at angle i_p, and a mutually perpendicular refracted ray at angle r.
Interpret: Demonstrates the 90-degree spatial separation between the reflected and refracted rays, confirming the geometric foundation of Brewster's law.
Exam use: Numerical problem visualization and conceptual ray tracing.
Pile of Plates Experiment
Draw: Several parallel glass plates arranged such that unpolarized light is incident at Brewster's angle, producing a transmitted beam that is strongly polarized parallel to the plane of incidence.
Interpret: Successive reflections remove the perpendicular polarization component, polarizing the transmitted beam through multiple partial reflections.
Exam use: Conceptual questions on generating polarized light without polaroids.
Solved numericals
TGT core
Calculate the polarizing angle for glass of refractive index 1.5. (Given tan(56.3 degrees) = 1.5)
Identify the given refractive index: mu = 1.5.
Use Brewster's law formula: tan($i_p) = \mu$.
Substitute the value: tan($i_p) = 1.5$.
Find the inverse tangent: $i_p = arctan(1.5) = 56.3 degrees$.
Answer: 56.3 degrees
PGT extension
Light is incident from air onto a water surface of refractive index 4/3 at the polarizing angle. Find the angle of refraction.
State the relationship at the polarizing angle for air-water interface: i_p + r = 90 degrees.
Calculate the angle of refraction r: $r = 90 degrees - 53.13 degrees$.
Perform subtraction: r = 36.87 degrees.
Answer: 36.87 degrees
PGT extension
The angle between the reflected and refracted rays of light when incident on a transparent medium is 90 degrees. If the angle of refraction is 30 degrees, find the refractive index of the medium.
Given that the angle between reflected and refracted rays is 90 degrees, the incidence angle i_p satisfies the Brewster condition: i_p + r = 90 degrees.
Apply Brewster's law to find refractive index: m$u = \tan(i_p) = \tan(60 degrees)$.
Evaluate tan(60 degrees) = sqrt(3) approx 1.732.
Answer: 1.732 (or sqrt(3))
Assumptions and limitations
Assumes the media are optically isotropic, non-magnetic (relative permeability mu_r approx 1), and transparent dielectrics.
Does not hold strictly for absorbing media (metals), where the refractive index is complex and minimum reflection does not yield complete polarization.
Common misconceptions
Common error: The reflected light is partially polarized only at angles greater than Brewster's angle.
Correct: The reflected light is partially polarized at all angles of incidence except normal incidence, but it is 100% plane-polarized exclusively at Brewster's angle.
Polarization degree varies continuously with angle, peaking at i_p.
Common error: At Brewster's angle, both the reflected and refracted rays are completely polarized.
Correct: Only the reflected ray is completely polarized. The refracted ray remains partially polarized because it contains a mixture of both parallel and perpendicular polarization components.
Refraction transmits some of both orthogonal electric field components.
Common error: Brewster's law applies only when light travels from air to a denser medium.
Correct: Brewster's law is valid for light passing between any two transparent dielectric media, where tan($i_p) = \mu_2 / \mu_1$.
The relative refractive index governs the relation regardless of whether mu_1 is unity.
Quick revision
Brewster's law formula is tan($i_p) = \mu_2 / \mu_1$, defining the angle where reflected light is completely plane-polarized.
Brewster's Law: tan($i_p) = \frac{\mu_2}{\mu_1}$
Brewster's Law for Air-Medium Interface: tan($i_p) = \mu$
Must remember
Brewster's law states that the tangent of the polarizing angle equals the relative refractive index of the medium: $\tan(i_p) = \mu_{21}$.
At Brewster's angle, the reflected and refracted rays are mutually perpendicular, meaning $i_p + r = 90^\circ$.
The reflected light at Brewster's angle is 100% plane-polarized with electric field oscillations perpendicular to the plane of incidence.
The refracted ray is only partially polarized because it retains both parallel and perpendicular polarization components.
Brewster's law is applicable to any transparent dielectric interface, not just air-to-medium boundaries.
Glossary
Polarizing Angle ($i_p$)The angle of incidence at which reflected light is completely plane-polarized parallel to the reflecting surface.
Brewster's LawThe physical law stating that $\tan(i_p) = \mu$, linking the refractive index to the polarizing angle.
Plane of IncidenceThe plane containing the incident ray, the normal to the interface, the reflected ray, and the refracted ray.
Relative Refractive Index ($\mu_{21}$)The ratio of the refractive index of medium 2 to that of medium 1, expressed as $\mu_2 / \mu_1$.
Reflected Ray vs Refracted Ray at Brewster's Angle
Feature
Reflected Ray
Refracted Ray
State of Polarization
Completely polarized (perpendicular to plane of incidence)
Partially polarized
Intensity
Fraction of incident unpolarized light intensity
Remaining transmitted intensity
Direction Condition
Angle of reflection equals $i_p$
Angle of refraction satisfies $r = 90^\circ - i_p$
Practice Quiz
Choose an option to reveal the explanation.
1. What is the angle between the reflected and refracted rays when unpolarized light is incident on a transparent medium at Brewster's angle?
Explanation: At Brewster's angle, the geometric condition requires that $i_p + r = 90^\circ$. Since the angle of reflection equals $i_p$, the angle between the reflected ray and refracted ray is $180^\circ - (i_p + r) = 180^\circ - 90^\circ = 90^\circ$.
2. Brewster's law is mathematically expressed as:
Explanation: Brewster discovered that the tangent of the polarizing angle of incidence equals the refractive index of the medium, yielding $\tan(i_p) = \mu$.
3. If the refractive index of a glass medium is $\sqrt{3}$, what is the polarizing angle for light incident from air?
4. At the polarizing angle, the state of polarization of the reflected ray is:
Explanation: At Brewster's angle, the electric field component parallel to the plane of incidence is completely transmitted into the medium, leaving the reflected ray entirely plane-polarized perpendicular to the plane of incidence.
5. Which of the following statements regarding the refracted ray at Brewster's angle is correct?
Explanation: The refracted ray is only partially polarized because it transmits both parallel and perpendicular components of the incident unpolarized light.
6. If the angle of refraction at Brewster's angle is $35^\circ$, what is the polarizing angle of incidence?
Explanation: The geometric condition at Brewster's angle is $i_p + r = 90^\circ$. Thus, $i_p = 90^\circ - 35^\circ = 55^\circ$.
7. Brewster's law for light traveling from medium 1 to medium 2 with refractive indices $\mu_1$ and $\mu_2$ is given by:
Explanation: The general form of Brewster's law is $\tan(i_p) = \mu_{21} = \frac{\mu_2}{\mu_1}$, representing the refractive index of the second medium with respect to the first.
8. Polarization by reflection depends primarily on:
Explanation: Brewster's law $\tan(i_p) = \mu_2 / \mu_1$ indicates that the polarizing angle depends exclusively on the refractive indices of the media forming the interface.
9. What happens to the reflected light from a dielectric surface when the angle of incidence is neither zero nor the Brewster angle?
Explanation: The reflected beam is partially polarized for all angles of incidence except normal incidence and the specific polarizing angle where it becomes 100% polarized.
10. If the refractive index of water is $4/3$, the polarizing angle is given by:
Explanation: From Brewster's law for an air-medium interface, $\tan(i_p) = \mu$, which gives $i_p = \arctan(\mu) = \arctan(4/3)$.
11. At Brewster's angle, the vibrations of the electric vector in the reflected ray are:
Explanation: The parallel component is entirely refracted at $i_p$, leaving only the electric field vector oscillating perpendicular to the plane of incidence in the reflected ray.
12. Which optical device utilizes Brewster's angle principles to minimize reflection losses for a specific polarization state?
Explanation: Brewster's windows are cut at Brewster's angle and used inside laser cavities to minimize reflection losses for the polarization parallel to the plane of incidence.
13. If light travels from a denser medium ($\mu_1$) into a rarer medium ($\mu_2$), can a Brewster angle exist?
Explanation: When light goes from denser to rarer, $\mu_{21} < 1$, which mathematically yields an angle whose tangent is less than 1 ($i_p = \arctan(\mu_2/\mu_1)$), but practically external reflection conditions apply.
14. In terms of phase velocity ($v_1$ and $v_2$), Brewster's law for an interface between two media can be expressed as:
Explanation: Since refractive index is inversely proportional to phase velocity ($\mu = c/v$), the relative refractive index $\mu_{21} = \mu_2 / \mu_1 = v_1 / v_2$, leading to $\tan(i_p) = v_1 / v_2$.
15. What is the primary reason why ordinary glass reflects unpolarized light partially rather than totally at angles other than $i_p$?
Explanation: Fresnel's equations show that reflection coefficients for parallel and perpendicular polarizations differ, leading to partial polarization at all angles except where the parallel reflection coefficient vanishes ($i_p$).
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. The polarizing angle for a piece of glass is $60^\circ$. What is the refractive index of the glass?
Explanation: According to Brewster's law, $\mu = \tan(i_p) = \tan(60^\circ) = \sqrt{3}$.
2. When light of a certain wavelength is incident on a transparent surface at the polarizing angle, the angle between the reflected and refracted rays is:
Explanation: At Brewster's angle, the reflected and refracted rays are mutually perpendicular, making the angle between them exactly $90^\circ$.
3. If the refractive index of a medium is $1.732$, the Brewster angle for light incident from air is approximately:
Explanation: Since $1.732 \approx \sqrt{3}$, the polarizing angle $i_p = \arctan(\sqrt{3}) = 60^\circ$.
4. At the polarizing angle, the reflected ray consists of light whose electric vector vibrates:
Explanation: At Brewster's angle, only the component perpendicular to the plane of incidence is reflected, making the reflected ray completely polarized with vibrations perpendicular to the plane of incidence.
5. The refractive index of water is $4/3$ and that of glass is $3/2$. What is the tangent of the polarizing angle when light travels from water to glass?
Explanation: For light traveling from medium 1 (water) to medium 2 (glass), $\tan(i_p) = \mu_2 / \mu_1 = (3/2) / (4/3) = 9/8$.
6. If the angle of incidence for a transparent slab is equal to the polarizing angle, the angle of refraction $r$ is related to $i_p$ by:
Explanation: The geometric condition for Brewster's angle is $i_p + r = 90^\circ$, which gives $r = 90^\circ - i_p$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. A ray of light is incident on the surface of a glass plate of refractive index $1.5$ at the polarizing angle. The angle of refraction is:
Explanation: At Brewster's angle, $i_p = \arctan(\mu) = \arctan(1.5)$. Since $r = 90^\circ - i_p$, the angle of refraction is $90^\circ - \arctan(1.5)$.
2. Which of the following conditions is essential for Brewster's law to be applicable?
Explanation: Brewster's law is valid for transparent, dielectric, non-magnetic, and isotropic media where reflection and refraction occur without significant absorption.
3. If the polarizing angle of a medium is $53^\circ$ ($ \tan 53^\circ \approx 1.327 $), its refractive index is approximately:
Explanation: By Brewster's law, $\mu = \tan(53^\circ) \approx 1.327$, which corresponds closely to water ($1.33$).
4. In electromagnetic wave boundary conditions, Brewster's angle corresponds to the angle of incidence where the reflection coefficient for which polarization vanishes?
Explanation: For TM (p-polarized) light, the electric field vector lies in the plane of incidence, and at Brewster's angle, its reflection coefficient becomes zero.
5. The phase change experienced by the reflected light when incident from a rarer medium onto a denser medium at Brewster's angle is:
Explanation: Upon reflection from an optically denser medium, light undergoes a phase change of $\pi$ ($180^\circ$) due to the boundary conditions of electric fields.
6. If a beam of unpolarized light is incident on a glass-air interface from the glass side at the polarizing angle, what is the value of $\tan(i_p)$?
Explanation: When light travels from glass ($\mu_1 = \mu$) to air ($\mu_2 = 1$), $\tan(i_p) = \mu_2 / \mu_1 = 1 / \mu$.
7. What is the speed of light in a medium whose polarizing angle from air is $60^\circ$? (Take $c = 3 \times 10^8$ m/s)
Explanation: Here $\mu = \tan(60^\circ) = \sqrt{3}$. The speed in the medium is $v = c / \mu = 3 \times 10^8 / \sqrt{3} = \sqrt{3} \times 10^8$ m/s.
8. Why is the transmitted (refracted) beam at Brewster's angle described as only partially polarized?
Explanation: Only the parallel component of reflection vanishes; however, the refracted beam contains contributions from both parallel and perpendicular incident components, making it partially polarized.
9. If the angle of incidence is varied from $0^\circ$ to $90^\circ$, the degree of polarization of the reflected light reaches $100\%$:
Explanation: Complete ($100\%$) plane-polarization of the reflected beam occurs solely at the specific incidence angle known as Brewster's angle.
10. A transparent dielectric block has a refractive index $\mu$. A ray of light is incident at Brewster's angle $i_p$. Which relation connects the angle of refraction $r$ and $\mu$?
Explanation: Since $i_p + r = 90^\circ$, $\tan(i_p) = \tan(90^\circ - r) = \cot(r) = \mu$. Thus, $\tan(r) = 1/\mu$.