MHT CET · Physics · Wave Optics
In Young's double slit experiment, the fifth maximum with wavelength ' \(\lambda_1\) ' is at a distance ' \(\mathrm{y}_1\) ' and the same maximum with wavelength ' \(\lambda_2\) ' is at a distance ' \(y_2\) ' measured from the central bright band. Then \(\frac{y_1}{y_2}\) is equal to [D and \(d\) are constant]
- A \(\frac{\lambda_1}{\lambda_2}\)
- B \(\frac{\lambda_2}{\lambda_1}\)
- C \(\frac{\lambda_1^2}{\lambda_2^2}\)
- D \(\frac{\lambda_2^2}{\lambda_1^2}\)
Answer & Solution
Correct Answer
(A) \(\frac{\lambda_1}{\lambda_2}\)
Step-by-step Solution
Detailed explanation
The equations for the position of the fringe from the central maxima are given as
\(\begin{aligned}
\mathrm{y}_1 & =\frac{5 \lambda_1 \mathrm{D}}{\mathrm{d}} \\
\mathrm{y}_2 & =\frac{5 \lambda_2 \mathrm{D}}{\mathrm{d}} \\
\therefore \quad \frac{\mathrm{y}_1}{\mathrm{y}_2} & =\frac{\lambda_1}{\lambda_2}
\end{aligned}\)
\(\begin{aligned}
\mathrm{y}_1 & =\frac{5 \lambda_1 \mathrm{D}}{\mathrm{d}} \\
\mathrm{y}_2 & =\frac{5 \lambda_2 \mathrm{D}}{\mathrm{d}} \\
\therefore \quad \frac{\mathrm{y}_1}{\mathrm{y}_2} & =\frac{\lambda_1}{\lambda_2}
\end{aligned}\)
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