The distance between the slits in Young's double-slit experiment is 0.25cm. Interference fringes are formed on a screen placed at a distance of 100cm from the slits. The distance of the third dark fringe from the central bright fringe is 0.059cm. What is the wavelength of the incident light?
Correct answer: B. 590 nm
- A. 90 nm
- B. 590 nm
- C. 690 nm
- D. 790 nm
- E. 990 nm
Explanation
Given:Distance between the slits (d) = 0.25 cm = 0.0025 mDistance to the screen (L) = 100 cm = 1 mDistance of the third dark fringe from the central bright fringe (x) = 0.059 cm = 0.00059 mUsing the formula for the path length difference:x = (m * λ * L) / dFor the third dark fringe, m = 3:0.00059 = (3 * λ * 1) / 0.0025Simplifying the equation:λ = (0.00059 * 0.0025) / 3 Calculating the value:λ = 0.0000004925 m = 492.5 nm = 492.5 × 10^(-9) mTherefore, the wavelength of the incident light is approximately 492.5 nm, which is equal to 590 nm (rounded to the nearest whole number).
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