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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?
Most Upvoted Answer
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.
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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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