WBJEE · Physics · Wave Optics
In Young's double slit experiment, light of wavelength \(\lambda\) passes through the double-slit and forms interference fringes on a screen \(1.2 \mathrm{~m}\) away. If the difference between 3 rd order maximum and 3 rd order minimum is \(0.18 \mathrm{~cm}\) and the slits are \(0.02 \mathrm{~cm}\) apart, then \(\lambda\) is
- A \(1200 \mathrm{~nm}\)
- B \(450 \mathrm{~nm}\)
- C \(600 \mathrm{~nm}\)
- D \(300 \mathrm{~nm}\)
Answer & Solution
Correct Answer
(C) \(600 \mathrm{~nm}\)
Step-by-step Solution
Detailed explanation
Distance between 3 rd order minima and 3 rd order maxima is \(\frac{\beta}{2}\) \( \begin{array}{l} \frac{\beta}{2}=0.18 \mathrm{~cm} \\ \beta=0.36 \mathrm{~cm} \\ \beta=\frac{\mathrm{D} \lambda}{\mathrm{d}} \\ \therefore \lambda=600 \mathrm{~nm} \end{array} \)
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