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An electromagnetic wave of frequency 3 MHz passes from vacuum into a medium with dielectric constant k = 4. Then
- a)both wavelength and frequency remain unchanged
- b)wavelength is doubled and frequency becomes half
- c)wavelength is halved and frequency remains unchanged
- d)wavelength is doubled and the frequency remains unchanged
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
An electromagnetic wave of frequency 3 MHz passes from vacuum into a m...
Explanation:
When an electromagnetic wave passes from vacuum into a medium with a dielectric constant, the speed of light changes. The speed of light in a medium is given by the equation:
v = c/n
Where:
v = speed of light in the medium
c = speed of light in vacuum
n = refractive index of the medium
1. Speed of light:
The speed of light in vacuum is a fundamental constant and remains constant. Therefore, the speed of light in vacuum remains unchanged.
2. Wavelength:
The wavelength of an electromagnetic wave is given by the equation:
λ = c/f
Where:
λ = wavelength
c = speed of light
f = frequency
Since the speed of light in vacuum remains constant, the wavelength is inversely proportional to the frequency. Therefore, if the frequency remains unchanged, the wavelength will also remain unchanged.
3. Frequency:
The frequency of an electromagnetic wave remains unchanged when it passes from one medium to another. Therefore, the frequency remains constant.
4. Dielectric constant:
The dielectric constant of a medium, denoted by k, is a measure of how easily electric charges can move through the medium. It affects the permittivity of the medium, which in turn affects the speed of light in the medium.
When an electromagnetic wave passes from vacuum into a medium with a dielectric constant, the speed of light in the medium decreases. This means that the refractive index of the medium, given by the equation:
n = c/v
increases. Since the refractive index is directly proportional to the dielectric constant, a higher dielectric constant leads to a higher refractive index.
Conclusion:
In this case, as the electromagnetic wave passes from vacuum into a medium with a dielectric constant of 4, the speed of light in the medium decreases, resulting in an increase in the refractive index. However, since the frequency remains unchanged and the speed of light in vacuum remains constant, the wavelength of the wave is halved. Therefore, the correct answer is option 'C': the wavelength is halved and the frequency remains unchanged.
When an electromagnetic wave passes from vacuum into a medium with a dielectric constant, the speed of light changes. The speed of light in a medium is given by the equation:
v = c/n
Where:
v = speed of light in the medium
c = speed of light in vacuum
n = refractive index of the medium
1. Speed of light:
The speed of light in vacuum is a fundamental constant and remains constant. Therefore, the speed of light in vacuum remains unchanged.
2. Wavelength:
The wavelength of an electromagnetic wave is given by the equation:
λ = c/f
Where:
λ = wavelength
c = speed of light
f = frequency
Since the speed of light in vacuum remains constant, the wavelength is inversely proportional to the frequency. Therefore, if the frequency remains unchanged, the wavelength will also remain unchanged.
3. Frequency:
The frequency of an electromagnetic wave remains unchanged when it passes from one medium to another. Therefore, the frequency remains constant.
4. Dielectric constant:
The dielectric constant of a medium, denoted by k, is a measure of how easily electric charges can move through the medium. It affects the permittivity of the medium, which in turn affects the speed of light in the medium.
When an electromagnetic wave passes from vacuum into a medium with a dielectric constant, the speed of light in the medium decreases. This means that the refractive index of the medium, given by the equation:
n = c/v
increases. Since the refractive index is directly proportional to the dielectric constant, a higher dielectric constant leads to a higher refractive index.
Conclusion:
In this case, as the electromagnetic wave passes from vacuum into a medium with a dielectric constant of 4, the speed of light in the medium decreases, resulting in an increase in the refractive index. However, since the frequency remains unchanged and the speed of light in vacuum remains constant, the wavelength of the wave is halved. Therefore, the correct answer is option 'C': the wavelength is halved and the frequency remains unchanged.
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An electromagnetic wave of frequency 3 MHz passes from vacuum into a medium with dielectric constant k = 4. Thena)both wavelength and frequency remain unchangedb)wavelength is doubled and frequency becomes halfc)wavelength is halved and frequency remains unchangedd)wavelength is doubled and the frequency remains unchangedCorrect answer is option 'C'. Can you explain this answer?
Question Description
An electromagnetic wave of frequency 3 MHz passes from vacuum into a medium with dielectric constant k = 4. Thena)both wavelength and frequency remain unchangedb)wavelength is doubled and frequency becomes halfc)wavelength is halved and frequency remains unchangedd)wavelength is doubled and the frequency remains unchangedCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An electromagnetic wave of frequency 3 MHz passes from vacuum into a medium with dielectric constant k = 4. Thena)both wavelength and frequency remain unchangedb)wavelength is doubled and frequency becomes halfc)wavelength is halved and frequency remains unchangedd)wavelength is doubled and the frequency remains unchangedCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electromagnetic wave of frequency 3 MHz passes from vacuum into a medium with dielectric constant k = 4. Thena)both wavelength and frequency remain unchangedb)wavelength is doubled and frequency becomes halfc)wavelength is halved and frequency remains unchangedd)wavelength is doubled and the frequency remains unchangedCorrect answer is option 'C'. Can you explain this answer?.
An electromagnetic wave of frequency 3 MHz passes from vacuum into a medium with dielectric constant k = 4. Thena)both wavelength and frequency remain unchangedb)wavelength is doubled and frequency becomes halfc)wavelength is halved and frequency remains unchangedd)wavelength is doubled and the frequency remains unchangedCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An electromagnetic wave of frequency 3 MHz passes from vacuum into a medium with dielectric constant k = 4. Thena)both wavelength and frequency remain unchangedb)wavelength is doubled and frequency becomes halfc)wavelength is halved and frequency remains unchangedd)wavelength is doubled and the frequency remains unchangedCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electromagnetic wave of frequency 3 MHz passes from vacuum into a medium with dielectric constant k = 4. Thena)both wavelength and frequency remain unchangedb)wavelength is doubled and frequency becomes halfc)wavelength is halved and frequency remains unchangedd)wavelength is doubled and the frequency remains unchangedCorrect answer is option 'C'. Can you explain this answer?.
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