Zero Reflection For P-Polarised Wave and Brewster Angle MCQ Quiz - Objective Question with Answer for Zero Reflection For P-Polarised Wave and Brewster Angle - Download Free PDF

Last updated on Jun 23, 2025

Latest Zero Reflection For P-Polarised Wave and Brewster Angle MCQ Objective Questions

Zero Reflection For P-Polarised Wave and Brewster Angle Question 1:

An unpolarized light beam travelling in air is incident on a medium of refractive index 1.73 at Brewster’s angle. Then:

  1. reflected light is completely polarized and the angle of reflection is close to 60°
  2. reflected light is partially polarized and the angle of reflection is close to 30°
  3. both reflected and transmitted light are perfectly polarized with angles of reflection and refraction close to 60° and 30°, respectively
  4. transmitted light is completely polarized with angle of refraction close to 30°

Answer (Detailed Solution Below)

Option 1 : reflected light is completely polarized and the angle of reflection is close to 60°

Zero Reflection For P-Polarised Wave and Brewster Angle Question 1 Detailed Solution

Correct option is : (1) Reflected light is completely polarized and the angle of reflection is close to 60°

Using Brewster's law

μ = tan θp

⇒ 1.73 = tan θp

⇒ √3 = tan θp

⇒ θp = 60°

1 (6)

Zero Reflection For P-Polarised Wave and Brewster Angle Question 2:

Light is incident on an interface between water (µ=4/3) and glass (µ=3/2). For total internal reflection, light should be travelling from:

  1. water to glass and \(\angle i > \angle i_c\)
  2. water to glass and \(\angle i < \angle i_c\)
  3. glass to water and \(\angle i < \angle i_c\)
  4. glass to water and \(\angle i > \angle i_c\)

Answer (Detailed Solution Below)

Option 4 : glass to water and \(\angle i > \angle i_c\)

Zero Reflection For P-Polarised Wave and Brewster Angle Question 2 Detailed Solution

The correct answer is - glass to water and ∠ i > ∠ i_c

Key Points

  • Total Internal Reflection
    • Total internal reflection occurs when light travels from a medium with a higher refractive index (glass) to a medium with a lower refractive index (water).
    • The critical angle (i_c) is the angle of incidence in the denser medium at which the refracted ray emerges along the interface.
    • For total internal reflection to occur, the angle of incidence (i) must be greater than the critical angle (i_c).

Additional Information

  • Refractive Index
    • The refractive index (μ) is a measure of how much the speed of light is reduced inside a medium.
    • In this context, glass has a refractive index of 3/2 and water has a refractive index of 4/3.
  • Critical Angle Calculation
    • The critical angle can be calculated using the formula: sin(i_c) = μ2 / μ1, where μ1 is the refractive index of the denser medium (glass) and μ2 is the refractive index of the less dense medium (water).
    • Here, sin(i_c) = (4/3) / (3/2) = 8/9, giving a critical angle i_c for the glass to water interface.
  • Angle of Incidence
    • If the angle of incidence i is greater than i_c, total internal reflection occurs, and light is reflected entirely within the denser medium (glass).
    • This phenomenon is utilized in optical fibers and other applications requiring efficient light transmission without loss.

Zero Reflection For P-Polarised Wave and Brewster Angle Question 3:

A wave pulse travelling on a two piece string gets partially reflected and partially transmitted at the junction. The reflected wave is inverted in shape as compared to the incident one. If the incident wave has wavelength λ and the transmitted wave λ-

  1. λ > λ
  2. λ′ = λ
  3. λ < λ
  4. nothing can be said about the relation of λ and λ'

Answer (Detailed Solution Below)

Option 3 : λ < λ

Zero Reflection For P-Polarised Wave and Brewster Angle Question 3 Detailed Solution

Calculation:

When a wave pulse travels on a two-piece string and reaches the junction, part of the wave is reflected and part is transmitted. The fact that the reflected wave is inverted indicates that the wave is encountering a denser medium at the junction.

The wavelength of a wave is given by the relation:

λ = v/f

Where:

  • λ is the wavelength,
  • v is the velocity of the wave,
  • f is the frequency of the wave.

When the wave travels from a less dense medium to a denser medium, the velocity of the wave decreases. Since the frequency of the wave remains constant, the wavelength of the wave in the denser medium will decrease as well.

Therefore, we have:

λ' < λ

Where:

  • λ is the wavelength of the incident wave,
  • λ' is the wavelength of the transmitted wave.

Final Answer: The correct answer is option 3.

Zero Reflection For P-Polarised Wave and Brewster Angle Question 4:

The critical angle of a crystal is 30°. Its Brewster angle is _______ degrees (Round off to the nearest integer).

Answer (Detailed Solution Below) 63

Zero Reflection For P-Polarised Wave and Brewster Angle Question 4 Detailed Solution

Explanation:

The critical angle (\( \theta_c \)) is related to the refractive index ( n ) as:
   
\( \sin \theta_c = \frac{1}{n}. \)
   Substituting\( \theta_c = 30^\circ\) :
    \( \sin 30^\circ = \frac{1}{n} \implies n = 2\).
   

Brewster's angle (\( \theta_B\) ) is given by:
   
\( \tan \theta_B = n\).
   
   Substituting n = 2 :
   
  \( \theta_B = \tan^{-1}(2)\).
   

Calculating:
    \( \theta_B \approx 63.4^\circ \text{ (rounded to 63°)}\).
   

Final Answer: \(\theta_B = 63^\circ\) .

Zero Reflection For P-Polarised Wave and Brewster Angle Question 5:

From the Brewster’s law of polarization it can be said that the polarization angle depends on 

  1. Wavelength of light
  2. Refractive index of the medium
  3. Rotation of plane of polarization
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 4 : More than one of the above

Zero Reflection For P-Polarised Wave and Brewster Angle Question 5 Detailed Solution

Concept:

Brewster's Law of Polarization:

Brewster's law states that the tangent of the angle of polarization is equal to the refractive index of the medium.

This relationship is given as μ = tan(θp), where μ is the refractive index and θp is the polarizing angle.

The refractive index μ can also be expressed in terms of the wavelength of light, as the refractive index depends on the wavelength.

Therefore, the polarization angle is influenced by both the refractive index and the wavelength of light.

Calculation:

According to Brewster's law:

μ = tan(θp), where θp is the angle of polarization.

Refractive index (μ) depends on the wavelength of light (λ) as it varies for different wavelengths in different media.

Hence, the polarization angle depends on:

Wavelength of light - Since μ varies with wavelength, the polarization angle changes with different wavelengths.

Refractive index of the medium - The angle of polarization depends directly on the refractive index of the medium.

Rotation of the plane of polarization - This also affects the polarization angle.

Therefore, the polarization angle depends on more than one factor:

∴ The correct option is: More than one of the above.

Top Zero Reflection For P-Polarised Wave and Brewster Angle MCQ Objective Questions

A uniform plane wave traveling in free space and having the electric field.

\({\rm{\vec E}} = \left( {\sqrt 2 {{{\rm{\hat a}}}_{\rm{x}}} - {{{\rm{\hat a}}}_{\rm{z}}}} \right)\cos \left[ {6\sqrt 3 {\rm{\pi }} \times {{10}^8}{\rm{t}} - 2{\rm{\pi }}\left( {{\rm{x}} + \sqrt 2 {\rm{z}}} \right)} \right]\frac{{\rm{V}}}{{\rm{m}}}\)

is incident on a dielectric medium (relative permittivity > 1, relative permeability = 1) as shown in the figure and there is no reflected wave.

GATE EC 2018 Solutions 10 May 2019 Rishi Madhu images Q16

The relative permittivity (correct to two decimal places) of the dielectric medium is

___________.

Answer (Detailed Solution Below) 1.9 - 2.1

Zero Reflection For P-Polarised Wave and Brewster Angle Question 6 Detailed Solution

Download Solution PDF

Concept:

GATE EC 2018 Solutions 10 May 2019 Rishi Madhu images Q16a

  the electric field shown in figure can be represented by 

\(\vec E = \left( {{E_x}{{\hat a}_x} + {E_z}{{\hat a}_z}} \right)\cos \left( {\omega t - \vec k.\vec r} \right)\)

Application:

GATE EC 2018 Solutions 10 May 2019 Rishi Madhu images Q16a

The given electric field is of the form

\(\vec E = \left( {{E_x}{{\hat a}_x} + {E_z}{{\hat a}_z}} \right)\cos \left( {\omega t - \vec k.\vec r} \right)\)

Where,

\(\vec k.\vec r = 2\pi \left( {x + \sqrt 2 z} \right)\)

\(\vec r\) is the position vector given by

\(\vec r = x{a_{\hat x}} + y{a_{\hat y}} + z{a_{\hat z}}\)

Thus, the propagation vector is given by:

\(\vec k = 2\pi x\;a\hat x + 2\pi \sqrt 2 \;a\hat z\)

Propagation vectors have x and z components, electric field also has x and z component.

This is the case of parallel polarization.

For parallel polarization

Brewster angle gives the angle where no-reflection occurs

\(\tan {\theta _i} = \sqrt {\frac{{{\epsilon_{{r_2}}}}}{{{\epsilon_{{r_1}}}}}} \)

Where,

\(\tan {\theta _i} = \frac{{{k_i}z}}{{{k_i}x}} = \frac{{2\pi \sqrt 2 }}{{2\pi }} = \sqrt 2 \)

Substitute value of tan θi

\(\sqrt 2 = \sqrt {\frac{{{\epsilon_{{r_2}}}}}{1}}\)

\({\epsilon_{{r_2}}} = 2\)

For parallel polarisation, for lossless dielectrics, the expression for Brewster angle for a wave traveling from medium 1 to the medium of refractive indices η1, and η2 respectively is

  1. \(\rm \theta =\sin^{-1}\left( \sqrt{\frac{\in_1/\in_2}{1+\sqrt{\frac{\in_1}{\in_2}}}} \right)\)
  2. \(\rm \theta =\sin^{-1}\left( \sqrt{\frac{\in_2/\in_1}{1+\in_2/\in_1}} \right)\)
  3. \(\rm \theta =\tan^{-1}\left( \sqrt{\frac{\in_2}{\in_1}} \right)\)
  4. \(\rm \theta =\cos^{-1}\left( \sqrt{\frac{\in_1}{\in_2}} \right)\)

Answer (Detailed Solution Below)

Option 3 : \(\rm \theta =\tan^{-1}\left( \sqrt{\frac{\in_2}{\in_1}} \right)\)

Zero Reflection For P-Polarised Wave and Brewster Angle Question 7 Detailed Solution

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Concept:

The Brewster law states the relationship between light waves and polarized light. The polarized light vanishes at this maximum angle.

Brewster angle is the angle when a wave is an incident on the surface of a perfect dielectric at which there is no reflected wave and the incident wave is parallelly polarised. 

For an elliptically polarized wave incident on the interface of a dielectric, the reflected wave will be elliptically polarized 

Brewster angle is given by:

\(\theta = {\tan ^{ - 1}}\sqrt {\frac{{{\epsilon_2}}}{{{\epsilon_1}}}}\)

F1 Tapesh 9.12.20 Pallavi D14

plane wave is generated under water (ϵ = 81 ϵ0 and μ = μ0) The wave is parallel polarized. At the interface between water and air, the angle a for which there is no reflection is

  1. 83.88° 
  2. 83.66° 
  3. 84.86°
  4. 84.08° 

Answer (Detailed Solution Below)

Option 2 : 83.66° 

Zero Reflection For P-Polarised Wave and Brewster Angle Question 8 Detailed Solution

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Concept:

Brewster’s Angle:  Brewster's angle is an angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection.

\(tan{\theta _i} = \sqrt {\frac{{{ϵ_2}}}{{{ϵ_1}}}} - - - \left( 1 \right)\)

Calculation:

For no reflection incident wave should be completely transmitted to another medium at Brewster angle.

Given:

ϵ2 = ϵ0, ϵ1 = 81ϵ0 

From (1) we get,

\(tan{\theta _i} = \sqrt {\frac{{{\epsilon_0}}}{{81{\epsilon_0}}}} \)

\({\theta _i} = {\tan ^{ - 1}}\frac{1}{9} = {6.34^o}\)

α = 90 – θi

α = 90 – 6.34o

α = 83.66o

Note: Brewster angle does not affect S polarized waves.

For no reflection condition, a vertically polarized wave should be incident at the interface between two dielectrics having ϵ1 = 4 and ϵ2 = 9, with an incident angle of 

  1. \(tan^{-1}(\frac{9}{4})\)
  2. \(tan^{-1}(\frac{3}{2})\)
  3. \(tan^{-1}(\frac{2}{3})\)
  4. \(tan^{-1}(\frac{4}{9})\)

Answer (Detailed Solution Below)

Option 2 : \(tan^{-1}(\frac{3}{2})\)

Zero Reflection For P-Polarised Wave and Brewster Angle Question 9 Detailed Solution

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Concept:

Brewster’s Angle:  Brewster's angle is an angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection.

\(tan{\theta _i} = \sqrt {\frac{{{\epsilon_2}}}{{{\epsilon_1}}}} \)----(1)

Calculation:

For no reflection incident wave should be completely transmitted to another medium at Brewster angle.

Given:

ϵ2 = 9, ϵ1 = 4

From equation (1) we get,

\(tan{\theta _i} = \sqrt {\frac{9}{4}} \)

\({\theta _i} = {\tan ^{ - 1}}\frac{3}{2}\)

Note: Brewster angle does not affect S polarized waves.

An unpolarized light beam travelling in air is incident on a medium of refractive index 1.73 at Brewster’s angle. Then:

  1. reflected light is completely polarized and the angle of reflection is close to 60°
  2. reflected light is partially polarized and the angle of reflection is close to 30°
  3. both reflected and transmitted light are perfectly polarized with angles of reflection and refraction close to 60° and 30°, respectively
  4. transmitted light is completely polarized with angle of refraction close to 30°

Answer (Detailed Solution Below)

Option 1 : reflected light is completely polarized and the angle of reflection is close to 60°

Zero Reflection For P-Polarised Wave and Brewster Angle Question 10 Detailed Solution

Download Solution PDF

Correct option is : (1) Reflected light is completely polarized and the angle of reflection is close to 60°

Using Brewster's law

μ = tan θp

⇒ 1.73 = tan θp

⇒ √3 = tan θp

⇒ θp = 60°

1 (6)

Zero Reflection For P-Polarised Wave and Brewster Angle Question 11:

An unpolarised light beam strikes a glass surface at Brewster's angle. Then 

  1. The reflected light will be partially polarised.
  2. The refracted light will be completely polarised. 
  3. Both the reflected and refracted light will be completely polarised.
  4. The reflected light will be completely polarised but the refracted light will be partially polarised.

Answer (Detailed Solution Below)

Option 4 : The reflected light will be completely polarised but the refracted light will be partially polarised.

Zero Reflection For P-Polarised Wave and Brewster Angle Question 11 Detailed Solution

Calculation:

F1 Savita UG Entrance 13-8-24 D4

According to Brewster's law, reflected rays are completely polarized and refracted rays are partially polarized.

Zero Reflection For P-Polarised Wave and Brewster Angle Question 12:

Brewster’s law in terms of refractive index μ and ip, where ip is the angle of incidence at which maximum polarization is observed, is expressed as:

  1. μ = cot ip
  2. μ = sin ip
  3. μ = cos ip
  4. μ = tan ip

Answer (Detailed Solution Below)

Option 4 : μ = tan ip

Zero Reflection For P-Polarised Wave and Brewster Angle Question 12 Detailed Solution

Concept:

  • The angle of incidence at which a beam of un-polarized light falling on a transparent surface is reflected as a beam of completely plane polarised light is called polarising or Brewster angle. It is denoted by ip.
  • Brewster's law: It states that when a ray is passed through some transparent medium having refractive index μ at any particular angle of incidence, reflected ray is completely polarized; and the angle between reflected and refracted ray is 900.

 

F2 J.K 19.5.20 Pallavi D5

μ = tan θB 

Where μ = refractive index and θB is Brewster's angle or polarizing angle (ip).

Explanation:

  • From above it is clear that the Brewster law states the relationship between light waves and polarized light
  • The correct option is μ = tan ip

Zero Reflection For P-Polarised Wave and Brewster Angle Question 13:

A uniform plane wave traveling in free space and having the electric field.

\({\rm{\vec E}} = \left( {\sqrt 2 {{{\rm{\hat a}}}_{\rm{x}}} - {{{\rm{\hat a}}}_{\rm{z}}}} \right)\cos \left[ {6\sqrt 3 {\rm{\pi }} \times {{10}^8}{\rm{t}} - 2{\rm{\pi }}\left( {{\rm{x}} + \sqrt 2 {\rm{z}}} \right)} \right]\frac{{\rm{V}}}{{\rm{m}}}\)

is incident on a dielectric medium (relative permittivity > 1, relative permeability = 1) as shown in the figure and there is no reflected wave.

GATE EC 2018 Solutions 10 May 2019 Rishi Madhu images Q16

The relative permittivity (correct to two decimal places) of the dielectric medium is

___________.

Answer (Detailed Solution Below) 1.9 - 2.1

Zero Reflection For P-Polarised Wave and Brewster Angle Question 13 Detailed Solution

Concept:

GATE EC 2018 Solutions 10 May 2019 Rishi Madhu images Q16a

  the electric field shown in figure can be represented by 

\(\vec E = \left( {{E_x}{{\hat a}_x} + {E_z}{{\hat a}_z}} \right)\cos \left( {\omega t - \vec k.\vec r} \right)\)

Application:

GATE EC 2018 Solutions 10 May 2019 Rishi Madhu images Q16a

The given electric field is of the form

\(\vec E = \left( {{E_x}{{\hat a}_x} + {E_z}{{\hat a}_z}} \right)\cos \left( {\omega t - \vec k.\vec r} \right)\)

Where,

\(\vec k.\vec r = 2\pi \left( {x + \sqrt 2 z} \right)\)

\(\vec r\) is the position vector given by

\(\vec r = x{a_{\hat x}} + y{a_{\hat y}} + z{a_{\hat z}}\)

Thus, the propagation vector is given by:

\(\vec k = 2\pi x\;a\hat x + 2\pi \sqrt 2 \;a\hat z\)

Propagation vectors have x and z components, electric field also has x and z component.

This is the case of parallel polarization.

For parallel polarization

Brewster angle gives the angle where no-reflection occurs

\(\tan {\theta _i} = \sqrt {\frac{{{\epsilon_{{r_2}}}}}{{{\epsilon_{{r_1}}}}}} \)

Where,

\(\tan {\theta _i} = \frac{{{k_i}z}}{{{k_i}x}} = \frac{{2\pi \sqrt 2 }}{{2\pi }} = \sqrt 2 \)

Substitute value of tan θi

\(\sqrt 2 = \sqrt {\frac{{{\epsilon_{{r_2}}}}}{1}}\)

\({\epsilon_{{r_2}}} = 2\)

Zero Reflection For P-Polarised Wave and Brewster Angle Question 14:

A wave which is parallel polarized and travelling in air (medium 1) is incident on a glass slab \({\epsilon_2} = 3.5\). If the wave is incident at Brewster’s angle then the angle (in degrees) between reflected and transmitted wave is(take condition just before it reaches Brewster's angle) ________.

Answer (Detailed Solution Below) 90

Zero Reflection For P-Polarised Wave and Brewster Angle Question 14 Detailed Solution

We have Brewster’s angle \({{\rm{\theta }}_{\rm{B}}} = {\tan ^{ - 1}}\left( {\sqrt {\frac{{{\epsilon_2}}}{{{\epsilon_1}}}} } \right) \Rightarrow {{\rm{\theta }}_{\rm{B}}} = {{\rm{\theta }}_1} = {\tan ^{ - 1}}\left( {\sqrt {\frac{{{\epsilon_2}}}{{{\epsilon_1}}}} } \right)\). Let angle of refraction be \({{\rm{\theta }}_2}\)

Then by, Snell’s law,

\(\frac{{{\rm{sin}}{{\rm{\theta }}_1}}}{{{\rm{sin}}{{\rm{\theta }}_2}}} = \sqrt {\frac{{{\epsilon_2}}}{{{\epsilon_1}}}}\)

Using, \(\tan {{\rm{\theta }}_1} = \sqrt {\frac{{{\epsilon_2}}}{{{\epsilon_1}}}}\) we have

\(\Rightarrow \sin {{\rm{\theta }}_1} = \frac{{\sqrt {{\epsilon_2}} }}{{\sqrt {{\epsilon_1} + {\epsilon_2}} }}\)

Thus, we have

\(\begin{array}{l} \sin {{\rm{\theta }}_2} = \sin {{\rm{\theta }}_1}\sqrt {\frac{{{\epsilon_1}}}{{{\epsilon_2}}}} \\ = \frac{{\sqrt {{\epsilon_2}} }}{{\sqrt {{\epsilon_1} + {\epsilon_2}} }}{\rm{}}\sqrt {\frac{{{\epsilon_1}}}{{{\epsilon_2}}}} \\ \Rightarrow \sin {{\rm{\theta }}_2} = \frac{{\sqrt {{\epsilon_1}} }}{{\sqrt {{\epsilon_1} + {\epsilon_2}} }} \end{array}\)

GSTE MOCK 2 Images-Q11

Now, angle between reflected and transmitted wave \({\rm{\theta }} = {180^ \circ }-\left( {{{\rm{\theta }}_1} + {{\rm{\theta }}_2}} \right)\)

\(\begin{array}{l} \Rightarrow {\rm{sin\theta }} = \sin \left( {180-\left( {{{\rm{\theta }}_1} + {{\rm{\theta }}_2}} \right)} \right) = \sin \left( {{{\rm{\theta }}_1} + {{\rm{\theta }}_2}} \right)\\ \Rightarrow \sin {\rm{\theta }} = \sin {{\rm{\theta }}_1}\cos {{\rm{\theta }}_2} + \sin {{\rm{\theta }}_2}\cos {{\rm{\theta }}_1}\\ = \frac{{\sqrt {{\epsilon_1}} }}{{\sqrt {{\epsilon_1} + {\epsilon_2}} }} \cdot \frac{{\sqrt {{\epsilon_1}} }}{{\sqrt {{\epsilon_1} + {\epsilon_2}} }} + \frac{{\sqrt {{\epsilon_2}} }}{{\sqrt {{\epsilon_1} + {\epsilon_2}} }} \cdot \frac{{\sqrt {{\epsilon_2}} }}{{\sqrt {{\epsilon_1} + {\epsilon_2}} }}\\ = \frac{{{\epsilon_1}}}{{{\epsilon_1} + {\epsilon_2}}} + \frac{{{\epsilon_2}}}{{{\epsilon_1} + {\epsilon_2}}} = \frac{{{\epsilon_1} + {\epsilon_2}}}{{{\epsilon_1} + {\epsilon_2}}}\\ \Rightarrow \sin {\rm{\theta }} = 1 \Rightarrow {\rm{\theta }} = 90^\circ \end{array}\)

Thus, when incidence is at Brewster’s angle, angle between reflected and transmitted wave is always 90°.

Zero Reflection For P-Polarised Wave and Brewster Angle Question 15:

A right circularly polarized (RCP) plane wave is incident at an angle of 60° to the

normal, on an air-dielectric interface. If the reflected wave is linearly polarized, the

relative dielectric constant \(\epsilon_{r2}\) is

Gate mobile 10          

  1. √2
  2. √3
  3. 2
  4. 3

Answer (Detailed Solution Below)

Option 4 : 3

Zero Reflection For P-Polarised Wave and Brewster Angle Question 15 Detailed Solution

Since, reflected wave is linearly polarized, angle of incidence is Brewster’s angle, \({\theta _B}\).

Now, brewster's angle \({\theta _B}\) is given by the relation

\( tan{\theta _B} = \sqrt {\frac{{\epsilon{_r}_2}}{{{\epsilon_r}_1}}} \Rightarrow tan60^\circ = \sqrt {\frac{{{\epsilon_r}_2}}{1}} \)

\(\Rightarrow {\epsilon_r}_2 = 3\)

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