JEE Mains · Physics · STD 12 - 10. Wave optics
In Young's double slit experiment the two slits are \(0.6 \; mm\) distance apart. Interference pattern is observed on a screen at a distance \(80 \; cm\) from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light will be \(\dots \; nm .\)
- A \(450\)
- B \(550\)
- C \(650\)
- D \(750\)
Answer & Solution
Correct Answer
(A) \(450\)
Step-by-step Solution
Detailed explanation
\(d=0.6 \times 10^{-3}\) \(D=80 \times 10^{-2}\) \(1st\) Dark fringe \(=\frac{D \lambda}{2 d}=\frac{d}{2}, \quad \lambda=\frac{d^{2}}{D}\) \(=450 \times 10^{-9} \; m\)
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