MHT CET · Physics · Wave Optics
In young's double slit experiment, the \(\mathrm{n}^{\text {th }}\) maximum of wavelength \(\lambda_1\) is at a distance of \(y_1\) from the central maximum. When the wavelength of the source is changed to \(\lambda_2,\left(\frac{\mathrm{n}}{3}\right)^{\text {th }}\) maximum is at a distance of \(y_2\) from its central maximum. The ratio \(\frac{y_1}{y_2}\) is
- A \(\frac{3 \lambda_1}{\lambda_2}\)
- B \(\frac{3 \lambda_2}{\lambda_1}\)
- C \(\frac{\lambda_1}{3 \lambda_2}\)
- D \(\frac{\lambda_2}{3 \lambda_1}\)
Answer & Solution
Correct Answer
(A) \(\frac{3 \lambda_1}{\lambda_2}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{y}_{\mathrm{n}}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{~d}}\)
For \(\mathrm{n}^{\text {th }}\) maximum
\(\mathrm{y}_1=\frac{\mathrm{n} \lambda_1 \mathrm{D}}{\mathrm{~d}}...(i)\)
For \(\left(\frac{\mathrm{n}}{3}\right)^{\mathrm{rd}}\) maximum
\(y_2=\frac{\frac{\mathrm{n}}{3} \lambda_2 \mathrm{D}}{\mathrm{~d}}...(ii)\)
\(\therefore \quad \frac{y_1}{y_2}=\frac{\frac{n \lambda_1 D}{d}}{\frac{\frac{n}{3} \lambda_2 D}{d}} \quad \ldots[\operatorname{From}(\mathrm{i})\) and (ii) \(]\)
\(\therefore \quad \frac{\mathrm{y}_1}{\mathrm{y}_2}=\frac{3 \lambda_1}{\lambda_2}\)
For \(\mathrm{n}^{\text {th }}\) maximum
\(\mathrm{y}_1=\frac{\mathrm{n} \lambda_1 \mathrm{D}}{\mathrm{~d}}...(i)\)
For \(\left(\frac{\mathrm{n}}{3}\right)^{\mathrm{rd}}\) maximum
\(y_2=\frac{\frac{\mathrm{n}}{3} \lambda_2 \mathrm{D}}{\mathrm{~d}}...(ii)\)
\(\therefore \quad \frac{y_1}{y_2}=\frac{\frac{n \lambda_1 D}{d}}{\frac{\frac{n}{3} \lambda_2 D}{d}} \quad \ldots[\operatorname{From}(\mathrm{i})\) and (ii) \(]\)
\(\therefore \quad \frac{\mathrm{y}_1}{\mathrm{y}_2}=\frac{3 \lambda_1}{\lambda_2}\)
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