Q.The polarising angle of a transparent medium is 30°. Calculate the angle of refraction. OR The diameter of objective lens of a Telescope is 1 meter. Calculate its resolving power for the light of wavelength 4538 Å.

When the reflected and refracted rays are perpendicular, the angle of incidence is the Brewster angle, and the refractive index equals its tangent.

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Brewster's law, $\tan\theta_B = n$, is the condition under which reflected light is completely (linearly) polarised — matching option (i).

Brewster's law states that when unpolarised light strikes a transparent medium at a particular angle (the polarising or Brewster angle $\theta_B$, where $\tan\theta_B = n$, the refractive index of the medium), the reflected ray is completely plane-polarised, wi …

The reflected and refracted rays are perpendicular exactly at the Brewster angle, where $\tan\theta_B=\mu$; for $\mu=1.5$ this is about $56.3^\circ$. In the alternative, the double-slit fringe-width formula gives $0.5$ mm.

Main part — Brewster's angle

When unpolarized light strikes a surface at the polarizing (Brewster) angle $\theta_B$, the reflected ray is completely (plane) polarized and the reflected and refracted rays are exactly perpendicular. Using Snell's law $\mu=\sin\theta_B/\sin\theta_r$ together with the geometric condition $\theta_B+\theta_r=90^\circ$ (so $\theta_r=90^\circ-\theta_B$, $\sin\theta_r=\cos\theta_B$):

$\mu = \frac{\sin\theta_B}{\cos\theta_B} = \tan\theta_B$

This is Brewster's law. With $\mu=1.5$:

$\theta_B = \tan^{-1}(1.5) \approx 56.3^\circ$

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At the Brewster angle, reflected light is completely plane-polarised, and $\tan\theta_B = n$.

Brewster's law states that when unpolarised light is incident on a transparent medium at a particular angle, called the Brewster angle or polarising angle $\theta_B$, the reflected light is completely plane-polarised (with its electric vector perpendicular to the plane of incidence). This angle satisfies $\tan\theta_B = n$, where $n$ is the refractive index of the medium on which light is incident (relative to the first medium). At this angle, the reflected ray and the re …

Brewster's angle occurs when reflected and refracted rays are perpendicular; for $\theta_B = \frac{\pi}{3}$, the refractive index is $\tan\theta_B$, giving $n = \sqrt{3}$.

$\tan\alpha=\mu$ and $\sin\beta=1/\mu$, so their product is 1 (option i).

Polarising (Brewster) angle. For light polarised on reflection, Brewster's law gives

$\mu=\tan\alpha.$

Critical angle. For total internal reflection,

$\sin\beta=\frac{1}{\mu}.$

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We use Brewster's Law, interpreting the polarizing angle to find the refractive index of the denser medium, and then use the definition of refractive index to calculate the speed of light. The speed of light in the denser medium is $\boxed{1.732 \times 10^8 \, \text{m/s}}$.

Reflected light is fully polarised at Brewster's (polarising) angle, defined by $\tan i_B = \mu$.

Concept. When unpolarised light reflects off a transparent surface, the reflected beam is partially polarised in general, but at one special angle it becomes completely plane-polarised (with vibrations perpendicular to the plane of incidence).

Brewster's law. At this angle the reflected and refracted rays are perpendicular to each other, giving …

tanα = μ (Brewster) and sinβ = 1/μ (critical angle) ⇒ cotα = sinβ: option (b).

By Brewster's law, at the polarising angle α the refractive index is

μ = tanα.

By the definition of the critical angle β,

sinβ = 1/μ.

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Brewster angle: tan(theta_B) = n = sqrt(3), giving theta_B = 60 degrees.

Brewster's law gives the polarising angle at which light reflected from a surface is completely plane-polarised:

$\tan\theta_B = n$

Step 1 — Substitute the refractive index: $\tan\theta_B = \sqrt{3}$.

Step 2 — $\theta_B = \tan^{-1}(\sqrt{3}) = 60^\circ$.

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Brewster's law gives $n=\tan\theta_p$, and $\tan 60° = \sqrt{3} \approx 1.732$.

When unpolarised light strikes the boundary of a transparent medium at a particular angle of incidence, called the polarising angle (or Brewster's angle) $\theta_p$, the reflected ray is found to be completely plane-polarised, with its electric vector oscillating perpendicular to the plane of incidence. Brewster discovered empirically (and it can be derived from the condition that the reflected and refracted rays are then exactly perpendicular to each other) that the refractive index of the medium is related to this angle by

$n = \tan\theta_p.$

This follows from combining Snell's law, $n = \dfrac{\sin\theta_p}{\sin\theta_r}$, with the geometric condition $\theta_p + \theta_r = 90°$ (reflected and refracted rays perpendicular), which gives $\sin\theta_r = \sin(90°-\theta_p)=\cos\theta_p$, so $n=\dfrac{\sin\theta_p}{\cos\theta_p}=\tan\theta_p$.

Substituting $\theta_p = 60°$:

$n = \tan 60° = \sqrt{3} \approx 1.732.$

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Brewster's law: $n = \tan\theta_p$, where $\theta_p$ is the polarizing (Brewster) angle.

When unpolarized light is incident on a transparent medium at a particular angle of incidence — the polarizing angle (or Brewster's angle) $\theta_p$ — the reflected light is found to be completely (linearly) polarized, with its plane of vibration perpendicular to the plane of incidence. At this angle, the reflected and refracted rays are perpendicular to each other ($\theta_p + r = 90^\circ$).

Applying Snell's law, $n = \dfrac{\sin\theta_p}{\sin r}$, and using $r = 90^\circ - \theta_p$ so $\sin r = \sin(90^\circ - \theta_p) = \cos\theta_p$:

$n = \dfrac{\sin\theta_p}{\cos\theta_p} = \tan\theta_p$

This relation is known as Brewster's law.

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