Plane Wave At Oblique Incidence MCQ Quiz - Objective Question with Answer for Plane Wave At Oblique Incidence - Download Free 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 θ_p

⇒ 1.73 = tan θ_p

⇒ √3 = tan θ_p

⇒ θ_p = 60°

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.

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.

But when it suffers total internal reflection, 100% reflection takes place with zero transmission.

So, in the above situation, till the angle of incidence is less than the critical angle, it will partially transmit and partially reflect. But after critical angle, all the light will reflect back.

Therefore, (C) is the correct option

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

when a transmission line is loaded with impedance it is represented as follows:

V = incident voltage

V’= reflected voltage

V’’= refracted or transmitted voltage.

Transmission line impedance is surge impedance Z_s.

Load impedance is Z_L.

Short circuited is line is considered when Z_L = 0

Coefficient of reflection is given by the expression \({V_{reflection}} = \frac{{V'}}{V}\)

\({V_{reflection}} = \frac{{{Z_L} - {Z_s}}}{{{Z_L} + {Z_S}}}\)

Calculations: 

\({{\rm{V}}_{{\rm{reflection}}}} = \frac{{0 - {Z_S}}}{{0 + {Z_S}}}\)

V _reflection = -1

Therefore coefficient of reflection of voltage for a short circuited line is -1.

CONCEPT:

• When light rays travel from one medium to another then there is bending in the light ray. This phenomenon is called as refraction of light.
• The refraction occurs due to different refractive indexes of the medium for different colours of light.
• When the light ray strikes on a small molecule of any medium particles then it gets spread in different directions then this phenomenon is called as a scattering of light.
• White light coming from the sun consists of seven constituent colours.
• These are – Violet, Indigo, blue, green, yellow, orange and red
• The red colour light has the longest wavelength among them and Violet has the shortest wavelength.

EXPLANATION:

• During the sunrise and sunset time, the light ray coming from the sun travel a longer distance in the atmosphere because the sun is on the horizon.
• The light having the least wavelength will be scattered most and light having the longest wavelength will be scattered least.
• Since the red and orange colour light has maximum wavelength among all other colours so it will be least scattered in the atmosphere and will come directly to our eye.
• As the combination of red and orange colours will come directly to our eyes so the sun appears reddish-orange. This is due to the least scattering by the atmosphere. Hence option 3 is correct.

Concept:

  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:

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 θ_i

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

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

Concept:

• Laws of reflection: The angle of reflection equals the angle of incidence.

• The incident ray, the reflected ray and the normal lie in the same plane.
• When light rays strikes on a smooth plane surface then the reflection is called regular reflection.
• When light reflects at many angles after the incidence on the surface, such a reflection is called as diffuse reflection.
• Light rays are not parallel to each other after irregular reflection, but they obeys laws of reflection at each point of reflection.

Explanation:

• In case of room, due to diffused reflection, the light gets spread in whole room and the room gets light. 
• The laws of reflection are followed by both- regular reflection as well as diffused reflection.
• The law of reflection is also valid for all curved and spherical surface

The given waveform can be represented as:

\(\vec E = \left( {8{{\hat a}_x} + 6{{\hat a}_y} + 5{{\hat a}_z}} \right){e^{j\left( {\omega t + 3x - 4y} \right)}}\)

A perfect conductor will reflect  \(\vec E\) completely

\({\vec E_t} = 6{\hat a_y} + 5{\hat a_z},\;{\vec E_n} = 8{\hat a_x}\)

For reflected wave \(\vec E\) tangential component will cancel out tangential component of \(\vec E\) so that net tangential field is zero.

i.e. \({\vec E_{{r_t}}} = - \left( {{{\vec E}_t}} \right) = - \left( {6{{\hat a}_y} + 5{{\hat a}_z}} \right)\)

Similarly for normal component, \({\vec E_{{r_n}}} = - {\vec E_n} = - \vec 8{a_x}\)

Since, the wave travelling is ‘-x’ & ‘+y’ direction and gets reflected at boundary x ≤ 0, the wave will travel along ‘+x’ & ‘+y’ direction.

\(\Rightarrow {\vec E_r} = \left( { - 8{{\hat a}_x} - 6{{\hat a}_y} - 5{{\hat a}_z}} \right){e^{j\left( {\omega t - 3x - 4y} \right)}}V/m\)

Note: It is an Official GATE Question, asked in GATE EC 2012 Paper. Please understand that Marks to All was allotted for this particular Question. This is because, in any TEM wave, Poynting Theorem must be satisfied according to which the propagation direction and field direction should be perpendicular.

If a light travels in a certain medium and it gets reflected off an optically denser medium with high refractive index, then it is regarded as External Reflection. 

By Snell’s law

\(\sqrt {{ϵ_1}} \sin {\theta _i} = \sqrt {{ϵ_2}} \sin {\theta _t}\)

\(1.\sin 45^\circ = \sqrt ϵ_r \sin {30}^\circ\)

\(\frac{2}{{\sqrt 2 }} = \sqrt ϵ_r\)

ϵ_r = 2

Critical Frequency :

The maximum frequency at which radio waves are totally reflected back from the ionosphere.

\(f_{c}=9 \sqrt{N_{max }}\)

fc  :  Critical Frequency

Nmax : Electron density

Usable frequency :

The range of frequency at which the propagation takes place for 50% of the month.

Cut-off Frequency :

The minimum frequency which is required for the propagation of waves.

Hence correct option is "2" 

The reflection coefficient (Γ) is defined as:

\({{\rm{\Gamma }}_E} = \frac{{{\eta _2} - {\eta _1}}}{{{\eta _2} + {\eta _1}}}\)

τ_E = 1 + Γ_E

Γ_H = - Γ_E

Application:

\(\frac{{{E_r}}}{{{E_i}}} = {{\rm{\Gamma }}_E} = \frac{{{\eta _2} - {\eta _1}}}{{{\eta _2} + {\eta _1}}}\)

\(= \frac{{{Z_2} - {Z_1}}}{{{Z_2} + {Z_1}}} = \frac{{{Z_c} - {Z_0}}}{{{Z_c} + {Z_0}}}\)

\({E_r} = \left( {\frac{{{z_c} - {z_0}}}{{{z_c} + {z_0}}}} \right){E_i}\)

τ = 1 + Γ

\(\tau = 1 + \;\frac{{{z_c} - {z_0}}}{{{z_c} + {z_0}}}\)

\(= \frac{{2{z_c}}}{{{z_c} + {z_0}}}\)

z_c = 0 (conductor impedance)

\(\tau = \frac{{2{z_c}}}{{{z_0}}}\)

\(\frac{{{E_t}}}{{{E_i}}} \simeq \frac{{2{z_c}}}{{{z_0}}}\)

\({E_i} \simeq \frac{{2{z_c}}}{{{z_0}}}{E_i}\)

Γ_H = - Γ_E

\({{\rm{\Gamma }}_H} = \frac{{{z_0} - {z_c}}}{{{z_0} + {z_c}}}\)

\({\tau _H} = 1 + {{\rm{\Gamma }}_{\rm{H}}} = 1 + \frac{{{z_0} - {z_c}}}{{{z_0} + {z_c}}}\)

\({\tau _H} = \frac{{2{z_0}}}{{{z_0} + {z_c}}}\)

For z_c = 0:

τ_H = 2

\(\frac{{{H_t}}}{{{H_i}}} = 2\)

H_t = 2H_i

The correct answer is 3 ft.

• To view an image of yourself in a plane mirror, you will need an amount of mirror equal to one-half of your height. A 6-foot tall man needs a 3-feet mirror (positioned properly) in order to view his entire image.

• If the top end of the mirror will be placed at the level of a person's eye then the image will be formed by following the law of reflection. 
• The angle of incidence = Angle of reflection.
• The part of the body in front of the mirror will be seen clearly by the eye. Moreover, a double-length will also get reflected from the end of the mirror.
• Thus if the total length of the body is L then we need the length of the mirror to be L/2 to see the complete image. So if the person's height is 6 feet, we need a minimum 3 feet mirror which is placed at the level of the eye to see the complete image of the person.
• The mirror must be at least half as tall as the person standing in front of it. Since the person is 180 cm tall, the vertical dimension of the mirror must be at least 90 cm. The lower edge of the mirror must be at a height that is half the distance between his feet and eyes.
• Since the eyes are located 6 cm from the top of his head which means it is at a height of 174 cm from his feet, the lower edge of the mirror must be 87 cm from the ground.
• When a ray of light is incident on a plane reflecting surface, the incident ray and the reflected ray will lie on the same plane. 
• The angle of incidence = Angle between the incident ray and the normal to the reflecting surface at the point of incidence.
• The Angle of reflection =  Angle between the reflected ray and the normal to the reflecting surface at the point of incidence.