AP EAMCET · PHYSICS · Wave Optics
In Young's double slit experiment, two slits are placed 2 mm from each other. Interference pattern is observed on a screen placed 2 m from the plane of the slits. Then the fringe width for a light of wavelength 400 nm is
- A \(0.4 \times 10^{-6} \mathrm{~m}\)
- B \(4 \times 10^{-6} \mathrm{~m}\)
- C \(0.4 \times 10^{-3} \mathrm{~m}\)
- D \(400 m\)
Answer & Solution
Correct Answer
(C) \(0.4 \times 10^{-3} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
In YDSE, \(\mathrm{d}=2 \mathrm{~mm}=2 \times 10^{-5} \mathrm{~m}, \mathrm{D}=2 \mathrm{~m}, \lambda=400 \mathrm{~nm}\) \(\therefore\) Fringe width, \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{400 \times 10^{-9} \times 2}{2 \times 10^{-3}}\)…
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