JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2021
- Hard
- MCQ
In Young's double slit arrangement, slits are separated by a gap of \(0.5\, mm ,\) and the screen is placed at a distance of \(0.5\, m\) from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of \(5890\, A^o\) is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(1178 \times 10^{-6}\,m\)
Step-by-step Solution
Detailed explanation
\(\beta=\frac{\lambda D}{d}=\frac{5890 \times 10^{-10} \times 0.5}{0.5 \times 10^{-3}}\)
\(=589 \times 10^{-6} m\)
Distance between first and third bright fringe is \(2 \beta=2 \times 589 \times 10^{-6} m\)
\(=1178 \times 10^{-6} m\)
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