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Consider a beam of plane polarised light of wavelength λ is incident on a optical component making angle 45° to the optical axis and output light is circularly polarised then the optical component is
- a)A full-wave plate
- b)A half-wave plate
- c)Q quarter wave plate
- d)An ordinary glass plate
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Consider a beam of plane polarised light of wavelength λis inci...
For quarter wave plate
n = 0,1,2, 3.......
When a plane polarised light is incident on quarter wave plate with angle θ= 45° then output light is circularly polarised.
n = 0,1,2, 3.......
When a plane polarised light is incident on quarter wave plate with angle θ= 45° then output light is circularly polarised.
Most Upvoted Answer
Consider a beam of plane polarised light of wavelength λis inci...
Λ and intensity I incident on a polarizer. The polarizer transmits only the component of the electric field vector that is parallel to its transmission axis. Let's assume that the transmission axis of the polarizer is aligned with the x-axis.
The electric field vector of the incident light can be written as E0*cos(ωt), where E0 is the amplitude of the electric field and ω is the angular frequency of the light.
The transmission axis of the polarizer is aligned with the x-axis, so it only transmits the component of the electric field vector that is parallel to the x-axis. Therefore, the transmitted electric field can be written as E0*cos(ωt) * cos(θ), where θ is the angle between the incident electric field vector and the x-axis.
The intensity of the transmitted light can be calculated using the formula I_transmitted = (1/2)*ε_0*c*E_transmitted^2, where ε_0 is the vacuum permittivity, c is the speed of light in vacuum, and E_transmitted is the amplitude of the transmitted electric field.
Substituting the expression for the transmitted electric field, we get:
I_transmitted = (1/2)*ε_0*c*(E0*cos(ωt) * cos(θ))^2
= (1/2)*ε_0*c*E0^2*cos^2(ωt) * cos^2(θ)
= (1/2)*ε_0*c*E0^2*cos^2(θ) * (1 + cos(2ωt))/2
= (1/4)*ε_0*c*E0^2*(1 + cos(2ωt))*cos^2(θ)
= (1/4)*ε_0*c*E0^2*cos^2(θ) + (1/4)*ε_0*c*E0^2*cos(2ωt)*cos^2(θ)
The term (1/4)*ε_0*c*E0^2*cos^2(θ) represents the average intensity of the transmitted light, which is given by:
I_avg = (1/4)*ε_0*c*E0^2*cos^2(θ)
So, the average intensity of the transmitted light is proportional to the square of the amplitude of the incident electric field (E0^2) and the square of the cosine of the angle between the incident electric field vector and the x-axis (cos^2(θ)).
The electric field vector of the incident light can be written as E0*cos(ωt), where E0 is the amplitude of the electric field and ω is the angular frequency of the light.
The transmission axis of the polarizer is aligned with the x-axis, so it only transmits the component of the electric field vector that is parallel to the x-axis. Therefore, the transmitted electric field can be written as E0*cos(ωt) * cos(θ), where θ is the angle between the incident electric field vector and the x-axis.
The intensity of the transmitted light can be calculated using the formula I_transmitted = (1/2)*ε_0*c*E_transmitted^2, where ε_0 is the vacuum permittivity, c is the speed of light in vacuum, and E_transmitted is the amplitude of the transmitted electric field.
Substituting the expression for the transmitted electric field, we get:
I_transmitted = (1/2)*ε_0*c*(E0*cos(ωt) * cos(θ))^2
= (1/2)*ε_0*c*E0^2*cos^2(ωt) * cos^2(θ)
= (1/2)*ε_0*c*E0^2*cos^2(θ) * (1 + cos(2ωt))/2
= (1/4)*ε_0*c*E0^2*(1 + cos(2ωt))*cos^2(θ)
= (1/4)*ε_0*c*E0^2*cos^2(θ) + (1/4)*ε_0*c*E0^2*cos(2ωt)*cos^2(θ)
The term (1/4)*ε_0*c*E0^2*cos^2(θ) represents the average intensity of the transmitted light, which is given by:
I_avg = (1/4)*ε_0*c*E0^2*cos^2(θ)
So, the average intensity of the transmitted light is proportional to the square of the amplitude of the incident electric field (E0^2) and the square of the cosine of the angle between the incident electric field vector and the x-axis (cos^2(θ)).
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Consider a beam of plane polarised light of wavelength λis incident on a optical component making angle 45° to the optical axis and output light is circularlypolarised then the optical component isa)A full-wave plateb)A half-wave platec)Q quarter wave plated)An ordinary glass plateCorrect answer is option 'C'. Can you explain this answer?
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Consider a beam of plane polarised light of wavelength λis incident on a optical component making angle 45° to the optical axis and output light is circularlypolarised then the optical component isa)A full-wave plateb)A half-wave platec)Q quarter wave plated)An ordinary glass plateCorrect answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Consider a beam of plane polarised light of wavelength λis incident on a optical component making angle 45° to the optical axis and output light is circularlypolarised then the optical component isa)A full-wave plateb)A half-wave platec)Q quarter wave plated)An ordinary glass plateCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a beam of plane polarised light of wavelength λis incident on a optical component making angle 45° to the optical axis and output light is circularlypolarised then the optical component isa)A full-wave plateb)A half-wave platec)Q quarter wave plated)An ordinary glass plateCorrect answer is option 'C'. Can you explain this answer?.
Consider a beam of plane polarised light of wavelength λis incident on a optical component making angle 45° to the optical axis and output light is circularlypolarised then the optical component isa)A full-wave plateb)A half-wave platec)Q quarter wave plated)An ordinary glass plateCorrect answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Consider a beam of plane polarised light of wavelength λis incident on a optical component making angle 45° to the optical axis and output light is circularlypolarised then the optical component isa)A full-wave plateb)A half-wave platec)Q quarter wave plated)An ordinary glass plateCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a beam of plane polarised light of wavelength λis incident on a optical component making angle 45° to the optical axis and output light is circularlypolarised then the optical component isa)A full-wave plateb)A half-wave platec)Q quarter wave plated)An ordinary glass plateCorrect answer is option 'C'. Can you explain this answer?.
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Consider a beam of plane polarised light of wavelength λis incident on a optical component making angle 45° to the optical axis and output light is circularlypolarised then the optical component isa)A full-wave plateb)A half-wave platec)Q quarter wave plated)An ordinary glass plateCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for Consider a beam of plane polarised light of wavelength λis incident on a optical component making angle 45° to the optical axis and output light is circularlypolarised then the optical component isa)A full-wave plateb)A half-wave platec)Q quarter wave plated)An ordinary glass plateCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of Consider a beam of plane polarised light of wavelength λis incident on a optical component making angle 45° to the optical axis and output light is circularlypolarised then the optical component isa)A full-wave plateb)A half-wave platec)Q quarter wave plated)An ordinary glass plateCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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