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An electromagnetic wave emitted by source travels 21km to arrive at a receiver the wave while traveling in another path is reflected from a surface at 19 km away and further travels 12Km to reach the same receiver .if destructive interference occurs at the receiving end,the maximum wavelength of the wave is 1) 0.5km 2) 1km 3) 5km 4) 10km?
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An electromagnetic wave emitted by source travels 21km to arrive at a ...
Given information:
- An electromagnetic wave travels from a source to a receiver.
- The wave travels a distance of 21 km to reach the receiver.
- The wave is reflected from a surface located 19 km away from the source.
- The reflected wave travels an additional 12 km to reach the receiver.
- Destructive interference occurs at the receiving end.
To find:
The maximum wavelength of the wave.
Explanation:
1. Understanding Destructive Interference:
Destructive interference occurs when two waves combine to form a resultant wave with a lower amplitude. It happens when the crests of one wave coincide with the troughs of the other wave, leading to a cancellation of amplitudes.
2. Path Difference:
In order for destructive interference to occur, the path difference between the two waves must be equal to an odd multiple of half the wavelength.
3. Path of the Waves:
In this scenario, the wave has two paths to reach the receiver:
- Path 1: Directly from the source to the receiver, a distance of 21 km.
- Path 2: Reflected from a surface located 19 km away and then travels an additional 12 km to reach the receiver.
4. Path Difference Calculation:
To calculate the path difference, we subtract the distance traveled by Path 2 from the distance traveled by Path 1:
Path Difference = Distance traveled by Path 1 - Distance traveled by Path 2
Path Difference = 21 km - (19 km + 12 km)
Path Difference = 21 km - 31 km
Path Difference = -10 km
5. Path Difference and Wavelength:
Since the path difference must be equal to an odd multiple of half the wavelength for destructive interference, we can write the following equation:
Path Difference = (2n + 1) * (λ/2)
Where n is an integer representing the number of half wavelengths.
6. Maximum Wavelength:
In order for the path difference to be -10 km, we can set n = -5 in the equation:
-10 km = (2(-5) + 1) * (λ/2)
-10 km = (-9) * (λ/2)
λ = (-10 km * 2) / -9
λ = 20 km / 9
λ ≈ 2.22 km
7. Comparing with Options:
The maximum wavelength of the wave is approximately 2.22 km, which is closest to option 3) 5 km.
Therefore, the maximum wavelength of the wave is 5 km.
- An electromagnetic wave travels from a source to a receiver.
- The wave travels a distance of 21 km to reach the receiver.
- The wave is reflected from a surface located 19 km away from the source.
- The reflected wave travels an additional 12 km to reach the receiver.
- Destructive interference occurs at the receiving end.
To find:
The maximum wavelength of the wave.
Explanation:
1. Understanding Destructive Interference:
Destructive interference occurs when two waves combine to form a resultant wave with a lower amplitude. It happens when the crests of one wave coincide with the troughs of the other wave, leading to a cancellation of amplitudes.
2. Path Difference:
In order for destructive interference to occur, the path difference between the two waves must be equal to an odd multiple of half the wavelength.
3. Path of the Waves:
In this scenario, the wave has two paths to reach the receiver:
- Path 1: Directly from the source to the receiver, a distance of 21 km.
- Path 2: Reflected from a surface located 19 km away and then travels an additional 12 km to reach the receiver.
4. Path Difference Calculation:
To calculate the path difference, we subtract the distance traveled by Path 2 from the distance traveled by Path 1:
Path Difference = Distance traveled by Path 1 - Distance traveled by Path 2
Path Difference = 21 km - (19 km + 12 km)
Path Difference = 21 km - 31 km
Path Difference = -10 km
5. Path Difference and Wavelength:
Since the path difference must be equal to an odd multiple of half the wavelength for destructive interference, we can write the following equation:
Path Difference = (2n + 1) * (λ/2)
Where n is an integer representing the number of half wavelengths.
6. Maximum Wavelength:
In order for the path difference to be -10 km, we can set n = -5 in the equation:
-10 km = (2(-5) + 1) * (λ/2)
-10 km = (-9) * (λ/2)
λ = (-10 km * 2) / -9
λ = 20 km / 9
λ ≈ 2.22 km
7. Comparing with Options:
The maximum wavelength of the wave is approximately 2.22 km, which is closest to option 3) 5 km.
Therefore, the maximum wavelength of the wave is 5 km.
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An electromagnetic wave emitted by source travels 21km to arrive at a receiver the wave while traveling in another path is reflected from a surface at 19 km away and further travels 12Km to reach the same receiver .if destructive interference occurs at the receiving end,the maximum wavelength of the wave is 1) 0.5km 2) 1km 3) 5km 4) 10km?
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An electromagnetic wave emitted by source travels 21km to arrive at a receiver the wave while traveling in another path is reflected from a surface at 19 km away and further travels 12Km to reach the same receiver .if destructive interference occurs at the receiving end,the maximum wavelength of the wave is 1) 0.5km 2) 1km 3) 5km 4) 10km? 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 emitted by source travels 21km to arrive at a receiver the wave while traveling in another path is reflected from a surface at 19 km away and further travels 12Km to reach the same receiver .if destructive interference occurs at the receiving end,the maximum wavelength of the wave is 1) 0.5km 2) 1km 3) 5km 4) 10km? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electromagnetic wave emitted by source travels 21km to arrive at a receiver the wave while traveling in another path is reflected from a surface at 19 km away and further travels 12Km to reach the same receiver .if destructive interference occurs at the receiving end,the maximum wavelength of the wave is 1) 0.5km 2) 1km 3) 5km 4) 10km?.
An electromagnetic wave emitted by source travels 21km to arrive at a receiver the wave while traveling in another path is reflected from a surface at 19 km away and further travels 12Km to reach the same receiver .if destructive interference occurs at the receiving end,the maximum wavelength of the wave is 1) 0.5km 2) 1km 3) 5km 4) 10km? 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 emitted by source travels 21km to arrive at a receiver the wave while traveling in another path is reflected from a surface at 19 km away and further travels 12Km to reach the same receiver .if destructive interference occurs at the receiving end,the maximum wavelength of the wave is 1) 0.5km 2) 1km 3) 5km 4) 10km? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electromagnetic wave emitted by source travels 21km to arrive at a receiver the wave while traveling in another path is reflected from a surface at 19 km away and further travels 12Km to reach the same receiver .if destructive interference occurs at the receiving end,the maximum wavelength of the wave is 1) 0.5km 2) 1km 3) 5km 4) 10km?.
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