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An electromagnetic wave of frequency v = 3.0MHz passes from vacuum into a dielectricmedium with relative permittivity ε = 4.0. Then[NEET Kar. 2013]
- a)wavelength is doubled and frequency isunchanged
- b)wavelength is doubled and frequencybecomes half
- c)wavelength is halved and frequencyremains unchanged
- d)wavelength and frequency both remainunchanged
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
An electromagnetic wave of frequency v = 3.0MHz passes from vacuum int...
Given: frequency f = 2MHz, relative
permittivity ∈r= 4
permittivity ∈r= 4
From formula,
velocity 
[Since frequency remains unchanged]
Most Upvoted Answer
An electromagnetic wave of frequency v = 3.0MHz passes from vacuum int...
Let's call the relative permittivity of the dielectric medium "εr".
The speed of light in vacuum is approximately 3.00 x 10^8 m/s.
The speed of light in the dielectric medium will be reduced by a factor of √εr, according to the formula:
v = c/√εr
where v is the speed of light in the dielectric medium, c is the speed of light in vacuum, and εr is the relative permittivity of the dielectric medium.
So, the speed of light in the dielectric medium will be:
v = (3.00 x 10^8 m/s) / √εr
The wavelength of the electromagnetic wave will also change as it passes into the dielectric medium. The formula for wavelength is:
λ = c/v
where λ is the wavelength, c is the speed of light in vacuum, and v is the speed of light in the dielectric medium.
So, the wavelength of the electromagnetic wave in the dielectric medium will be:
λ = (3.00 x 10^8 m/s) / [(3.00 x 10^8 m/s) / √εr]
Simplifying this expression:
λ = λ0 / √εr
where λ0 is the wavelength of the electromagnetic wave in vacuum.
Finally, we can calculate the wavelength of the electromagnetic wave in the dielectric medium, knowing its frequency:
v = fλ
where f is the frequency and λ is the wavelength.
So,
λ = v/f = (3.00 x 10^8 m/s) / (3.0 x 10^6 Hz) = 100 m
Therefore, the wavelength of the electromagnetic wave in vacuum is 100 meters.
Using the formula we derived earlier,
λ = λ0 / √εr
we can solve for λ0:
λ0 = λ x √εr = 100 m x √4 = 200 m
Therefore, the wavelength of the electromagnetic wave in the dielectric medium is 200 meters.
The speed of light in vacuum is approximately 3.00 x 10^8 m/s.
The speed of light in the dielectric medium will be reduced by a factor of √εr, according to the formula:
v = c/√εr
where v is the speed of light in the dielectric medium, c is the speed of light in vacuum, and εr is the relative permittivity of the dielectric medium.
So, the speed of light in the dielectric medium will be:
v = (3.00 x 10^8 m/s) / √εr
The wavelength of the electromagnetic wave will also change as it passes into the dielectric medium. The formula for wavelength is:
λ = c/v
where λ is the wavelength, c is the speed of light in vacuum, and v is the speed of light in the dielectric medium.
So, the wavelength of the electromagnetic wave in the dielectric medium will be:
λ = (3.00 x 10^8 m/s) / [(3.00 x 10^8 m/s) / √εr]
Simplifying this expression:
λ = λ0 / √εr
where λ0 is the wavelength of the electromagnetic wave in vacuum.
Finally, we can calculate the wavelength of the electromagnetic wave in the dielectric medium, knowing its frequency:
v = fλ
where f is the frequency and λ is the wavelength.
So,
λ = v/f = (3.00 x 10^8 m/s) / (3.0 x 10^6 Hz) = 100 m
Therefore, the wavelength of the electromagnetic wave in vacuum is 100 meters.
Using the formula we derived earlier,
λ = λ0 / √εr
we can solve for λ0:
λ0 = λ x √εr = 100 m x √4 = 200 m
Therefore, the wavelength of the electromagnetic wave in the dielectric medium is 200 meters.
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An electromagnetic wave of frequency v = 3.0MHz passes from vacuum into a dielectricmedium with relative permittivity ε = 4.0. Then[NEET Kar. 2013]a)wavelength is doubled and frequency isunchangedb)wavelength is doubled and frequencybecomes halfc)wavelength is halved and frequencyremains unchangedd)wavelength and frequency both remainunchangedCorrect answer is option 'C'. Can you explain this answer?
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An electromagnetic wave of frequency v = 3.0MHz passes from vacuum into a dielectricmedium with relative permittivity ε = 4.0. Then[NEET Kar. 2013]a)wavelength is doubled and frequency isunchangedb)wavelength is doubled and frequencybecomes halfc)wavelength is halved and frequencyremains unchangedd)wavelength and frequency both remainunchangedCorrect answer is option 'C'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about An electromagnetic wave of frequency v = 3.0MHz passes from vacuum into a dielectricmedium with relative permittivity ε = 4.0. Then[NEET Kar. 2013]a)wavelength is doubled and frequency isunchangedb)wavelength is doubled and frequencybecomes halfc)wavelength is halved and frequencyremains unchangedd)wavelength and frequency both remainunchangedCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electromagnetic wave of frequency v = 3.0MHz passes from vacuum into a dielectricmedium with relative permittivity ε = 4.0. Then[NEET Kar. 2013]a)wavelength is doubled and frequency isunchangedb)wavelength is doubled and frequencybecomes halfc)wavelength is halved and frequencyremains unchangedd)wavelength and frequency both remainunchangedCorrect answer is option 'C'. Can you explain this answer?.
An electromagnetic wave of frequency v = 3.0MHz passes from vacuum into a dielectricmedium with relative permittivity ε = 4.0. Then[NEET Kar. 2013]a)wavelength is doubled and frequency isunchangedb)wavelength is doubled and frequencybecomes halfc)wavelength is halved and frequencyremains unchangedd)wavelength and frequency both remainunchangedCorrect answer is option 'C'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about An electromagnetic wave of frequency v = 3.0MHz passes from vacuum into a dielectricmedium with relative permittivity ε = 4.0. Then[NEET Kar. 2013]a)wavelength is doubled and frequency isunchangedb)wavelength is doubled and frequencybecomes halfc)wavelength is halved and frequencyremains unchangedd)wavelength and frequency both remainunchangedCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electromagnetic wave of frequency v = 3.0MHz passes from vacuum into a dielectricmedium with relative permittivity ε = 4.0. Then[NEET Kar. 2013]a)wavelength is doubled and frequency isunchangedb)wavelength is doubled and frequencybecomes halfc)wavelength is halved and frequencyremains unchangedd)wavelength and frequency both remainunchangedCorrect answer is option 'C'. Can you explain this answer?.
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