MHT CET · Physics · Wave Optics
In Young's double slit experiment, \(8^{\text {th }}\) maximum with wavelength ' \(\lambda_1\) ' is at a distance ' \(\mathrm{d}_1\) ' from the central maximum and \(6^{\text {th }}\) maximum with wavelength ' \(\lambda_2\) ' is at a distance ' \(\mathrm{d}_2\) '. Then \(\frac{\mathrm{d}_2}{\mathrm{~d}_1}\) is
- A \(\frac{3 \lambda_1}{4 \lambda_2}\)
- B \(\frac{3 \lambda_2}{4 \lambda_1}\)
- C \(\frac{4 \lambda_1}{3 \lambda_2}\)
- D \(\frac{4 \lambda_2}{3 \lambda_1}\)
Answer & Solution
Correct Answer
(B) \(\frac{3 \lambda_2}{4 \lambda_1}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll} & \mathrm{d} \propto \mathrm{n} \lambda \\ \therefore \quad & \frac{\mathrm{d}_2}{\mathrm{~d}_1}=\frac{\mathrm{n}_2 \lambda_2}{\mathrm{n}_1 \lambda_1}=\frac{6 \lambda_2}{8 \lambda_1} \\ \therefore \quad & \frac{\mathrm{d}_2}{\mathrm{~d}_1}=\frac{3 \lambda_2}{4 \lambda_1}\end{array}\)
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