NEET UG 2020 — Physics Optics

2020mcqhard

The Brewsters angle ${i}_{b}$ for an interface should be

Official previous-year question

Held on 30 Apr 2020 · Verified 9 Jul 2026.

Options

A
${30}^{\text{o}}<{i}_{b}<{45}^{\text{o}}$
B
${45}^{\text{o}}<{i}_{b}<{90}^{\text{o}}$
C
${i}_{b}={90}^{\text{o}}$
D
${0}^{\text{o}}<{i}_{b}<{30}^{\text{o}}$

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Solution

Brewster's angle is defined as the angle of incidence at which the reflection of particular polarized light is zero, and it perfectly transmitted through a transparent surface.

Brewster's angle( ${i}_{b}$ ) is given by, $\mathrm{tan}({i}_{b})=\frac{{\mu }_{2}}{{\mu }_{1}}=\frac{{\mu }_{2}}{1}$ , where ${\mu }_{2}$ is the refractive index of the dielectric and ${\mu }_{1}$ is the refractive index of air.

We know that, ${\mu }_{2}>1$ . (refractive index of any dielectric will be greater than refractive index of air.)

$\Rightarrow \infty <\mathrm{tan}{i}_{b}>1$

$\Rightarrow {90}^{o}>{i}_{b}>{45}^{o}$

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Solution

Brewster's angle is defined as the angle of incidence at which the reflection of particular polarized light is zero, and it perfectly transmitted through a transparent surface.

We know that, ${\mu }_{2}>1$ . (refractive index of any dielectric will be greater than refractive index of air.)

$\Rightarrow \infty <\mathrm{tan}{i}_{b}>1$

$\Rightarrow {90}^{o}>{i}_{b}>{45}^{o}$

NEET UG 2020 — Physics Optics

Options

Did you get this right?

Solution

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NEET UG Optics in other years

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Other Physics topics for NEET UG

Mechanics

Electromagnetism

Modern Physics

Thermodynamics

Waves & Oscillations

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NEET UG 2020 — Physics Optics

Options

Did you get this right?

Solution

Your note

NEET UG Optics in other years

More NEET UG Optics questions

Other Physics topics for NEET UG

Mechanics

Electromagnetism

Modern Physics

Thermodynamics

Waves & Oscillations

Keep practicing Optics