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A single slit Fraunhofer diffraction pattern in 41. formed with white light. For what wavelength of light the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength 6500Å?
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A single slit Fraunhofer diffraction pattern in 41. formed with white ...
Calculating Wavelength of Light for Third Secondary Maximum
To determine the wavelength of light for which the third secondary maximum coincides with the second secondary maximum for red light of wavelength 6500Å, we can use the formula for the position of secondary maxima in single slit Fraunhofer diffraction:
d sinθ = mλ
where:
- d is the width of the slit
- θ is the angle of diffraction
- m is the order of the maximum
- λ is the wavelength of light
Finding the Position of the Third Secondary Maximum
For the third secondary maximum, m = 3. Let's assume the angle of diffraction for this maximum is θ3.
Therefore, we have:
d sinθ3 = 3λ
Comparison with Red Light Wavelength
Given that the red light has a wavelength of 6500Å, we can compare the positions of the third secondary maximum for the unknown wavelength (let's call it λ') and the second secondary maximum for red light:
d sinθ3 = 3λ'
d sinθ2 = 2(6500Å)
Equating the Two Expressions
Since the two maxima coincide, we have:
3λ' = 2(6500Å)
Solving for λ', we get:
λ' = 2(6500Å) / 3
Therefore, the wavelength of light for which the third secondary maximum coincides with the second secondary maximum in the diffraction pattern for red light of wavelength 6500Å is approximately 4333.33Å.
To determine the wavelength of light for which the third secondary maximum coincides with the second secondary maximum for red light of wavelength 6500Å, we can use the formula for the position of secondary maxima in single slit Fraunhofer diffraction:
d sinθ = mλ
where:
- d is the width of the slit
- θ is the angle of diffraction
- m is the order of the maximum
- λ is the wavelength of light
Finding the Position of the Third Secondary Maximum
For the third secondary maximum, m = 3. Let's assume the angle of diffraction for this maximum is θ3.
Therefore, we have:
d sinθ3 = 3λ
Comparison with Red Light Wavelength
Given that the red light has a wavelength of 6500Å, we can compare the positions of the third secondary maximum for the unknown wavelength (let's call it λ') and the second secondary maximum for red light:
d sinθ3 = 3λ'
d sinθ2 = 2(6500Å)
Equating the Two Expressions
Since the two maxima coincide, we have:
3λ' = 2(6500Å)
Solving for λ', we get:
λ' = 2(6500Å) / 3
Therefore, the wavelength of light for which the third secondary maximum coincides with the second secondary maximum in the diffraction pattern for red light of wavelength 6500Å is approximately 4333.33Å.
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A single slit Fraunhofer diffraction pattern in 41. formed with white light. For what wavelength of light the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength 6500Å?
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A single slit Fraunhofer diffraction pattern in 41. formed with white light. For what wavelength of light the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength 6500Å? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A single slit Fraunhofer diffraction pattern in 41. formed with white light. For what wavelength of light the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength 6500Å? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A single slit Fraunhofer diffraction pattern in 41. formed with white light. For what wavelength of light the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength 6500Å?.
A single slit Fraunhofer diffraction pattern in 41. formed with white light. For what wavelength of light the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength 6500Å? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A single slit Fraunhofer diffraction pattern in 41. formed with white light. For what wavelength of light the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength 6500Å? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A single slit Fraunhofer diffraction pattern in 41. formed with white light. For what wavelength of light the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength 6500Å?.
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