JEE Mains · Physics · STD 12 - 10. Wave optics
A fringe width of \(\,6 mm\) was produced for two slits separated by \(1\, mm\) apart. The screen is placed \(10\, m\) away. The wavelength of light used is \('x'\, nm.\) The value of \('x'\) to the nearest integer is
- A \(400\)
- B \(600\)
- C \(500\)
- D \(700\)
Answer & Solution
Correct Answer
(B) \(600\)
Step-by-step Solution
Detailed explanation
\(\beta=\frac{\lambda D }{ d }\) \(\lambda=\frac{\beta d }{ D }\) \(\lambda=\frac{6 \times 10^{-3} \times 10^{-3}}{10}\) \(\lambda=6 \times 10^{-7} m =600 \times 10^{-9} m\) \(\lambda=600 nm\)
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