JEE Mains · Physics · STD 12 - 10. Wave optics
White light is passed through a double slit and interference is observed on a screen \(1.5 \,{m}\) away. The separation between the slits is \(0.3 \,{mm}\). The first violet and red fringes are formed \(2.0 \,{mm}\) and \(3.5\, {mm}\) away from the central white fringes. The difference in wavelengths of red and voilet light is \(.... \,{nm} .\)
- A \(150\)
- B \(1300\)
- C \(300\)
- D \(600\)
Answer & Solution
Correct Answer
(C) \(300\)
Step-by-step Solution
Detailed explanation
Position of bright fringe \({y}={n} \frac{{D} \lambda}{{d}}\) \({y}_{1} \text { of red }=\frac{{D} \lambda_{{r}}}{{d}}=3.5\, {mm}\) \(\lambda_{{T}}=3.5 \times 10^{-3} \frac{{d}}{{D}}\) Similarly \(\lambda_{{v}}=2 \times 10^{-3} \frac{{d}}{{D}}\)…
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