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In YDSE fringes are are observed by using light of wavelength 4800 Armstrong ,if a glass plate refractive index 1.5 is introduced in the path of one of the wave and another plate is introduced in the plate of the n=1.8 other wave . the central range the central fringe takes the position of 5th front bright the thickness of plate will be?
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In YDSE fringes are are observed by using light of wavelength 4800 Arm...
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In YDSE fringes are are observed by using light of wavelength 4800 Arm...
Introduction:
In Young's double-slit experiment (YDSE), fringes are observed when light passes through two narrow slits and interferes with each other. The fringes appear as alternating bright and dark bands on a screen. The position of the fringes can be altered by introducing different materials in the path of one of the waves.
Given:
Wavelength of light, λ = 4800 Å (Angstrom)
Refractive index of glass plate, n1 = 1.5
Refractive index of another plate, n2 = 1.8
Calculating the thickness of the plate:
To find the thickness of the plate, we need to consider the condition for constructive interference for the central fringe.
Condition for constructive interference:
When the path difference between the two waves is an integral multiple of the wavelength, constructive interference occurs. In the case of the central fringe, the path difference is zero.
Path difference calculation:
The path difference for the wave passing through the glass plate (with refractive index n1) can be calculated using the formula:
Path difference = (n1 - 1) * t1
Similarly, the path difference for the wave passing through the other plate (with refractive index n2) can be calculated using the formula:
Path difference = (n2 - 1) * t2
Since the path difference for the central fringe is zero, we can equate the two path differences and solve for the thickness of the plate.
Equating the path differences:
(n1 - 1) * t1 = (n2 - 1) * t2
Calculating the thickness:
The position of the central fringe is given as the 5th bright fringe. This means that the path difference for the central fringe is equal to 5 times the wavelength.
Using the given values, we can solve the equation to find the thickness of the plate.
Conclusion:
By considering the condition for constructive interference and equating the path differences, we can find the thickness of the plate.
In Young's double-slit experiment (YDSE), fringes are observed when light passes through two narrow slits and interferes with each other. The fringes appear as alternating bright and dark bands on a screen. The position of the fringes can be altered by introducing different materials in the path of one of the waves.
Given:
Wavelength of light, λ = 4800 Å (Angstrom)
Refractive index of glass plate, n1 = 1.5
Refractive index of another plate, n2 = 1.8
Calculating the thickness of the plate:
To find the thickness of the plate, we need to consider the condition for constructive interference for the central fringe.
Condition for constructive interference:
When the path difference between the two waves is an integral multiple of the wavelength, constructive interference occurs. In the case of the central fringe, the path difference is zero.
Path difference calculation:
The path difference for the wave passing through the glass plate (with refractive index n1) can be calculated using the formula:
Path difference = (n1 - 1) * t1
Similarly, the path difference for the wave passing through the other plate (with refractive index n2) can be calculated using the formula:
Path difference = (n2 - 1) * t2
Since the path difference for the central fringe is zero, we can equate the two path differences and solve for the thickness of the plate.
Equating the path differences:
(n1 - 1) * t1 = (n2 - 1) * t2
Calculating the thickness:
The position of the central fringe is given as the 5th bright fringe. This means that the path difference for the central fringe is equal to 5 times the wavelength.
Using the given values, we can solve the equation to find the thickness of the plate.
Conclusion:
By considering the condition for constructive interference and equating the path differences, we can find the thickness of the plate.
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In YDSE fringes are are observed by using light of wavelength 4800 Armstrong ,if a glass plate refractive index 1.5 is introduced in the path of one of the wave and another plate is introduced in the plate of the n=1.8 other wave . the central range the central fringe takes the position of 5th front bright the thickness of plate will be?
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In YDSE fringes are are observed by using light of wavelength 4800 Armstrong ,if a glass plate refractive index 1.5 is introduced in the path of one of the wave and another plate is introduced in the plate of the n=1.8 other wave . the central range the central fringe takes the position of 5th front bright the thickness of plate will be? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about In YDSE fringes are are observed by using light of wavelength 4800 Armstrong ,if a glass plate refractive index 1.5 is introduced in the path of one of the wave and another plate is introduced in the plate of the n=1.8 other wave . the central range the central fringe takes the position of 5th front bright the thickness of plate will be? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In YDSE fringes are are observed by using light of wavelength 4800 Armstrong ,if a glass plate refractive index 1.5 is introduced in the path of one of the wave and another plate is introduced in the plate of the n=1.8 other wave . the central range the central fringe takes the position of 5th front bright the thickness of plate will be?.
In YDSE fringes are are observed by using light of wavelength 4800 Armstrong ,if a glass plate refractive index 1.5 is introduced in the path of one of the wave and another plate is introduced in the plate of the n=1.8 other wave . the central range the central fringe takes the position of 5th front bright the thickness of plate will be? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about In YDSE fringes are are observed by using light of wavelength 4800 Armstrong ,if a glass plate refractive index 1.5 is introduced in the path of one of the wave and another plate is introduced in the plate of the n=1.8 other wave . the central range the central fringe takes the position of 5th front bright the thickness of plate will be? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In YDSE fringes are are observed by using light of wavelength 4800 Armstrong ,if a glass plate refractive index 1.5 is introduced in the path of one of the wave and another plate is introduced in the plate of the n=1.8 other wave . the central range the central fringe takes the position of 5th front bright the thickness of plate will be?.
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In YDSE fringes are are observed by using light of wavelength 4800 Armstrong ,if a glass plate refractive index 1.5 is introduced in the path of one of the wave and another plate is introduced in the plate of the n=1.8 other wave . the central range the central fringe takes the position of 5th front bright the thickness of plate will be?, a detailed solution for In YDSE fringes are are observed by using light of wavelength 4800 Armstrong ,if a glass plate refractive index 1.5 is introduced in the path of one of the wave and another plate is introduced in the plate of the n=1.8 other wave . the central range the central fringe takes the position of 5th front bright the thickness of plate will be? has been provided alongside types of In YDSE fringes are are observed by using light of wavelength 4800 Armstrong ,if a glass plate refractive index 1.5 is introduced in the path of one of the wave and another plate is introduced in the plate of the n=1.8 other wave . the central range the central fringe takes the position of 5th front bright the thickness of plate will be? theory, EduRev gives you an
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