Practice question

Given: In a double-slit experiment, if lambda = 460 nm, d = 0.2 mm, and D = 2.0 m, what is the distance of the third bright fringe from the ntral maximum? These values define the system as per NCERT data. Formula: Bright fringe position x_n = n lambda D/d. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For the third bright fringe, n = 3 . lambda = 4.6 × 10⁻⁷ m, d = 2.0 × 10⁻⁴ m, D = 2.0 m . x_3 = frac3 × 4.6 × 10⁻⁷ × 2.02.0 × 10⁻⁴= 6.9 × 10⁻³ m = 6.9 mm . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.