In a Young's double slit experiment, the slits are 2 mm apart and are illuminated with a mixture of two wavelengths λ0 = 750nm and λ = 900nm. The minimum distance from the common central bright fringe on a screen 2m from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other is:
Correct answer: C. 4.5 mm
- A. 1.5 mm
- B. 3 mm
- C. 4.5 mm
- D. 6 mm
Explanation
In a Young's double slit experiment with two different wavelengths, the interference pattern is determined by the path length difference between the two waves. For the given scenario with λ0 = 750nm and λ = 900nm, the path length difference that leads to the coincidence of bright fringes from both interference patterns occurs at 4.5 mm from the central bright fringe on the screen. This distance is calculated based on the specific wavelengths and their corresponding phase differences.
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