JEE Mains · Physics · STD 12 - 10. Wave optics
In a Young's double slit experiment, slits are separated by \(0.5\ mm\), and the screen is placed \(150\ cm\) away. A beam of light consisting of two wavelengths, \(650\ nm\) and \(520\ nm\), is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is...... \(mm\)
- A \(1.56\)
- B \(7.8\)
- C \(9.75\)
- D \(15.6\)
Answer & Solution
Correct Answer
(B) \(7.8\)
Step-by-step Solution
Detailed explanation
For common maxima, \(n_{1} \lambda_{1}=n_{2} \lambda_{2}\) \(\Rightarrow \quad \frac{n_{1}}{n_{2}}=\frac{\lambda_{2}}{\lambda_{1}}=\frac{520 \times 10^{-9}}{650 \times 10^{-9}}=\frac{4}{5}\) For \(\lambda_{1}\) \(y=\frac{n_{1} \lambda_{1} D}{d}, \lambda_{1}=650\, \mathrm{nm}\)…
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