Unpolarised light passes from a rarer into a denser medium. If the reflected and the refracted rays are mutual

Question

Q.Unpolarised light passes from a rarer into a denser medium. If the reflected and the refracted rays are mutually perpendicular, the reflected light is linearly polarised ___________ to the plane of incidence.

CBSECBSE Class XII Board 2020Subjective· 1mImportance★★★★★est

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Polarisation by Reflection

Polarisation by Reflection – From Intuition to the Law

Imagine sunlight hitting the surface of a still lake. You see a glare — that harsh, bright reflection that makes it hard to see the fish below. Now put on polarised sunglasses and tilt your head. The glare vanishes. What just happened? The sunglasses blocked something that the reflection had done to the light.

That something is polarisation.

What does "polarisation" even mean?

Light is a transverse wave — the electric field oscillates perpendicular to the direction the light travels. In ordinary (unpolarised) light, the electric field vibrates in every possible direction perpendicular to the ray, all at once. Think of it like a skipping rope being shaken randomly in all sideways directions.

When light reflects off a surface, something interesting happens: the surface "filters" these vibrations. Certain directions of vibration get reflected more strongly than others. The reflected light is no longer vibrating in all directions — it is now partially polarised, and at one special angle, completely polarised.

The key intuition: the surface is picky

When light hits a boundary between two media (say, air and glass), the electrons in the glass are set into oscillation by the incoming electric field. These oscillating electrons then re-radiate light — that re-radiated light is the reflected beam.

But here is the crucial point: an oscillating electron cannot radiate along its own direction of oscillation. If the electron is vibrating up-and-down, it sends no energy straight along that up-down line.

Now, at a particular angle of incidence, the reflected ray and the refracted ray are perpendicular to each other. At that exact geometry, the direction of vibration that is parallel to the plane of incidence (call it the "p-polarisation") would require the electrons to radiate along their own oscillation direction — which they cannot do. So that component is completely suppressed. Only the vibration perpendicular to the plane of incidence (the "s-polarisation") survives.

The result: the reflected light is 100% plane-polarised, with its electric field vibrating perpendicular to the plane of incidence.

The precise statement

Important

When unpolarised light is reflected from a transparent surface (like glass or water), the reflected light is completely plane-polarised if the angle of incidence $\theta_i$ satisfies

$\tan \theta_B = \frac{n_2}{n_1}$

where $n_1$ is the refractive index of the incident medium and $n_2$ that of the transmitting medium. This angle $\theta_B$ is called Brewster's angle.

At Brewster's angle, the reflected and refracted rays are exactly $90^\circ$ apart. The reflected beam contains only the component of light whose electric field is perpendicular to the plane of incidence.

What about other angles? …

Start your 14-day free trial to unlock the full solution →

When reflected and refracted rays are perpendicular, the light is incident at Brewster's angle; the reflected light is then linearly polarised perpendicular to the plane of incidence.

When unpolarised light reflects off a dielectric interface, something remarkable happens at a special angle: the reflected beam becomes completely polarised. This occurs because of how the electric field components parallel and perpendicular to the plane of incidence behave differently during reflection.

The plane of incidence is the plane containing both the incident ray and the normal to the surface. Any light wave can be decomposed into two independent polarisation components: one with its electric field oscillating parallel to this plane (p-polarised) and one perpendicular to it (s-polarised, from the German senkrecht).

At most angles, both components are partially reflected. But at Brewster's angle $\theta_B$ , the p-polarised component is not reflected at all — it is entirely transmitted into the denser medium. Only the s-polarised component reflects, making the reflected beam completely linearly polarised.

The condition given in the problem — that reflected and refracted rays are mutually perpendicular — is precisely the geometric signature of Brewster's angle.

Here's why:

Set up the geometry. Let the angle of incidence be $\theta_i$ and the angle of refraction be $\theta_r$ . The reflected ray makes angle $\theta_i$ with the normal on the opposite side.
Apply the perpendicularity condition. If the reflected and refracted rays are perpendicular, then the angle between them is $90°$ . Looking at the geometry: the reflected ray is at angle $\theta_i$ from the normal (going back), and the refracted ray is at angle $\theta_r$ from the normal (going forward into the medium). For these to be perpendicular:

\theta_i + \theta_r = 90°

Connect to Snell's law. We have $n_1 \sin\theta_i = n_2 \sin\theta_r$ . Since $\theta_r = 90° - \theta_i$ , we get:

n_1 \sin\theta_i = n_2 \sin(90° - \theta_i) = n_2 \cos\theta_i

\tan\theta_i = \frac{n_2}{n_1}

This is Brewster's law, and $\theta_i = \theta_B$ is Brewster's angle. …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: For an unpolarized wave the displacement will be randomly changing with time though it will always be ______ to the direction of propagation.$
›Reveal solutionSolution

Light is transverse: the vibrations are always perpendicular to the direction of propagation.

Light is a transverse electromagnetic wave, meaning the electric field (displacement) oscillates in a direction perpendicular to the direction in which the wave travels. In unpolarized light, the direction of this vibration keeps changing randomly with time (in all directions …

Start your 14-day free trial to see the full solution →
CBSE 2020Set 55/2/11 mark
$Q.Unpolarised light passes from a rarer into a denser medium. If the reflected and the refracted rays are mutually perpendicular, the reflected light is linearly polarised ___________ to the plane of incidence.$
›Reveal solutionSolution
When reflected and refracted rays are perpendicular, the light is incident at Brewster's angle; the reflected light is then linearly polarised perpendicular to the plane of incidence.

When unpolarised light reflects off a dielectric interface, something remarkable happens at a special angle: the reflected beam becomes completely polarised. This occurs because of how the electric field components parallel and perpendicular to the plane of incidence behave differently during reflection.

The plane of incidence is the plane containing both the incident ray and the normal to the surface. Any light wave can be decomposed into two independent polarisation components: one with its electric field oscillating parallel to this plane (p-polarised) and one perpendicular to it (s-polarised, from the German senkrecht).

At most angles, both components are partially reflected. But at Brewster's angle $\theta_B$ , the p-polarised component is not reflected at all — it is entirely transmitted into the denser medium. Only the s-polarised component reflects, making the reflected beam completely linearly polarised.

The condition given in the problem — that reflected and refracted rays are mutually perpendicular — is precisely the geometric signature of Brewster's angle.

Here's why:

Set up the geometry. Let the angle of incidence be $\theta_i$ and the angle of refraction be $\theta_r$ . The reflected ray makes angle $\theta_i$ with the normal on the opposite side.

Apply the perpendicularity condition. If the reflected and refracted rays are perpendicular, then the angle between them is $90°$ . Looking at the geometry: the reflected ray is at angle $\theta_i$ from the normal (going back), and the refracted ray is at angle $\theta_r$ from the normal (going forward into the medium). For these to be perpendicular:

$\theta_i + \theta_r = 90°$

Connect to Snell's law. We have $n_1 \sin\theta_i = n_2 \sin\theta_r$ . Since $\theta_r = 90° - \theta_i$ , we get:

$n_1 \sin\theta_i = n_2 \sin(90° - \theta_i) = n_2 \cos\theta_i$ $\tan\theta_i = \frac{n_2}{n_1}$
This is Brewster's law, and $\theta_i = \theta_B$ is Brewster's angle. …
Start your 14-day free trial to see the full solution →

Q.Unpolarised light passes from a rarer into a denser medium. If the reflected and the refracted rays are mutually perpendicular, the reflected light is linearly polarised ___________ to the plane of incidence.

🎓Unlock everything free for 14 days

Unlock everything free for 14 days