A double-slit experiment with 0.20 mm slit separation and 5.0x102 nm light shows the first bright fringe at 3.0 mm. What is the distance to the screen?

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BlackTom AI · Accepted Answer

To begin, restate the problem setup and the options to ensure clarity: the slit separation is d = 0.20 mm = 2.0 × 10^-4 m, the light wavelength is λ = 5.0 × 10^2 nm = 5.0 × 10^-7 m, and the first bright fringe appears at a screen position y = 3.0 mm = 3.0 × 10^-3 m. The answer choices for the screen distance L are: a) 1.0 m, b) 0.90 m, c) 1.2 m, d) 1.1 m.

Option a: 1.0 m. Using the double-slit bright fringe condition d sin θ = m λ with m = 1 for the first bright fringe, we have sin θ = λ/d = (5.0 × 10^-7 m) / (2.0 × 10^-4 m) = 2.5 × 10^-3. For small angles, tan θ ≈ sin θ ≈ θ, and the diffraction geometry gives y ≈ L sin θ ≈ L tan θ. Solving for L yields L ≈ y / sin θ ≈ (3.0 × 10^-3 m) / (2.5 × 10^-3) ≈ 1.2 m. The given value of 1.0 m would require sin θ ≈ y/L = 3.0 × 10^-3 / 1.0 ≈ 3.0 × 10^-3, which is close but slightly larger than the actual sin θ; this discrepancy indicates 1.0 m is not consistent with the exact relation for the first bright fringe.

Option b: 0.90 m. Following the same approach, L ≈ y / sin θ = (3.0 × 10^-3 m) / (2.5 × 10^-3) ≈ 1.2 m. If you push L down to 0.90 m, sin θ would need to be y/L ≈ (3.0 × 10^-3) / (0.90) ≈ 3.33 × 10^-3, which is slightly larger than λ/d. This would imply a larger angle than allowed by the first bright fringe condition, so 0.90 m is inconsistent.

Option c: 1.2 m. This result comes directly from combining the first-bright-fringe condition with the small-angle geometry: sin θ = λ/d and y ≈ L sin θ, so L ≈ y d / λ. Substituting y = 3.0 × 10^-3 m, d = 2.0 × 10^-4 m, and λ = 5.0 × 10^-7 m gives L ≈ (3.0 × 10^-3)(2.0 × 10^-4) / (5.0 × 10^-7) = (6.0 × 10^-7) / (5.0 × 10^-7) = 1.2 m. This matches the standard derivation and the given data.

Option d: 1.1 m. If we compute from the same relation L ≈ y d / λ, we get 1.2 m, not 1.1 m. A screen distance of 1.1 m would correspond to sin θ = y / L ≈ (3.0 × 10^-3) / (1.1) ≈ 2.73 × 10^-3, which would yield m λ ≈ d sin θ ≈ (2.0 × 10^-4)(2.73 × 10^-3) ≈ 5.46 × 10^-7 m, slightly larger than λ, misplacing the first bright fringe.

Across the options, the calculation tied to the standard formula for double-slit interference with the first bright fringe aligns with L = y d / λ, yielding 1.2 m as the consistent screen distance. The other choices diverge from this relationship either by using an incompatible angle estimate or by misapplying the small-angle approximation, leading to values that do not satisfy d sin θ = λ for m = 1.

A double-slit experiment with 0.20 mm slit separation and 5.0x102 nm light shows the first bright fringe at 3.0 mm. What is the distance to the screen?单项选择题

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