CUET · PHYSICS · PYQ PAPER 2025
In a Young's double slit experiment, the distance between slits is 2.5 mm and the screen is placed 2 m away from the slits. The fringe width for a light of wavelength 400 nm is
- A 2.5 mm
- B 3.32 mm
- C 0.16 mm
- D 0.32 mm
Answer & Solution
Correct Answer
(D) 0.32 mm
Step-by-step Solution
Detailed explanation
\(\beta = \frac{\lambda D}{d}\) \(\beta = \frac{(400 \times 10^{-9} \text{ m}) \times (2 \text{ m})}{2.5 \times 10^{-3} \text{ m}}\) \(\beta = 0.32 \times 10^{-3} \text{ m} = 0.32 \text{ mm}\)
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