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In a Young’s double slit experiment with light of wavelength λ the separation of slits is d and distance of screen is D such that D >> d >> λ. If the fringe width is β, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is:
- a)β/6
- b)β/3
- c)β/4
- d)β/2
Correct answer is option 'C'. Can you explain this answer?
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In a Young’s double slit experiment with light of wavelength λ the se...
Given parameters:
- Wavelength of light: λ
- Separation of slits: d
- Distance of screen: D (D >> d >> λ)
- Fringe width: β
To find: Distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side
Formulae used:
- Fringe width: β = λD/d
- Path difference between two waves: δ = d sinθ
- Phase difference between two waves: Δφ = 2πδ/λ
- Intensity: I = I0 cos²(Δφ/2)
Calculation:
- The maximum intensity occurs at the center of the screen, where the path difference between the two waves is zero.
- Let's consider a point on one side of the maximum intensity point, where the path difference between two waves is δ. This point will be at an angle of θ = tan⁻¹(δ/D) with respect to the center.
- The phase difference between the two waves at this point is Δφ = 2πδ/λ.
- The intensity at this point is given by I = I0 cos²(Δφ/2).
- When the intensity falls to half of the maximum intensity, cos²(Δφ/2) = 1/2.
- Solving for Δφ, we get Δφ = π/2.
- Substituting the value of Δφ in the formula for δ, we get δ = λ/2 sinθ.
- For small angles, sinθ ≈ θ, so δ ≈ λ/2 (δ/D).
- This implies that the distance from the maximum intensity point to the point where intensity falls to half of maximum intensity on either side is given by δ/2, which is equal to β/4.
- Therefore, the correct option is (c) β/4.
Answer: Option (c) β/4.
- Wavelength of light: λ
- Separation of slits: d
- Distance of screen: D (D >> d >> λ)
- Fringe width: β
To find: Distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side
Formulae used:
- Fringe width: β = λD/d
- Path difference between two waves: δ = d sinθ
- Phase difference between two waves: Δφ = 2πδ/λ
- Intensity: I = I0 cos²(Δφ/2)
Calculation:
- The maximum intensity occurs at the center of the screen, where the path difference between the two waves is zero.
- Let's consider a point on one side of the maximum intensity point, where the path difference between two waves is δ. This point will be at an angle of θ = tan⁻¹(δ/D) with respect to the center.
- The phase difference between the two waves at this point is Δφ = 2πδ/λ.
- The intensity at this point is given by I = I0 cos²(Δφ/2).
- When the intensity falls to half of the maximum intensity, cos²(Δφ/2) = 1/2.
- Solving for Δφ, we get Δφ = π/2.
- Substituting the value of Δφ in the formula for δ, we get δ = λ/2 sinθ.
- For small angles, sinθ ≈ θ, so δ ≈ λ/2 (δ/D).
- This implies that the distance from the maximum intensity point to the point where intensity falls to half of maximum intensity on either side is given by δ/2, which is equal to β/4.
- Therefore, the correct option is (c) β/4.
Answer: Option (c) β/4.
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Community Answer
In a Young’s double slit experiment with light of wavelength λ the se...
Multiplying equation (i) and (ii) we get, y = β/4
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In a Young’s double slit experiment with light of wavelength λ the separation of slits is d and distance of screen is D such that D >> d >> λ. If the fringe width is β, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is:a)β/6b)β/3c)β/4d)β/2Correct answer is option 'C'. Can you explain this answer?
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In a Young’s double slit experiment with light of wavelength λ the separation of slits is d and distance of screen is D such that D >> d >> λ. If the fringe width is β, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is:a)β/6b)β/3c)β/4d)β/2Correct answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about In a Young’s double slit experiment with light of wavelength λ the separation of slits is d and distance of screen is D such that D >> d >> λ. If the fringe width is β, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is:a)β/6b)β/3c)β/4d)β/2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a Young’s double slit experiment with light of wavelength λ the separation of slits is d and distance of screen is D such that D >> d >> λ. If the fringe width is β, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is:a)β/6b)β/3c)β/4d)β/2Correct answer is option 'C'. Can you explain this answer?.
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