MHT CET · Physics · Wave Optics
In an interference experiment, the \(\mathrm{n}^{\text {th }}\) bright fringe for light of wavelength \(\lambda_1(n=0,1,2,3 \ldots)\) coincides with the \(\mathrm{m}^{\text {th }}\) dark fringe for light of wavelength \(\lambda_2(m=1,2,3 \ldots)\). The ratio \(\frac{\lambda_1}{\lambda_2}\) is
- A \(\frac{m-1}{n}\)
- B \(\frac{2 \mathrm{~m}-1}{\mathrm{n}}\)
- C \(\frac{2 \mathrm{~m}-1}{2 \mathrm{n}}\)
- D \(\frac{2 \mathrm{~m}+1}{2 \mathrm{n}}\)
Answer & Solution
Correct Answer
(C) \(\frac{2 \mathrm{~m}-1}{2 \mathrm{n}}\)
Step-by-step Solution
Detailed explanation
As \(\mathrm{n}^{\text {th }}\) bright fringe coincides with \(\mathrm{m}^{\text {th }}\) dark fringe
\(\begin{array}{ll}
\therefore & \frac{\mathrm{n} \lambda_1 \mathrm{D}}{\mathrm{~d}}=\frac{(2 \mathrm{~m}-1) \lambda_2 \mathrm{D}}{2 \mathrm{~d}} \\
\therefore & \frac{\lambda_1}{\lambda_2}=\frac{(2 \mathrm{~m}-1)}{2 \mathrm{n}}
\end{array}\)
\(\begin{array}{ll}
\therefore & \frac{\mathrm{n} \lambda_1 \mathrm{D}}{\mathrm{~d}}=\frac{(2 \mathrm{~m}-1) \lambda_2 \mathrm{D}}{2 \mathrm{~d}} \\
\therefore & \frac{\lambda_1}{\lambda_2}=\frac{(2 \mathrm{~m}-1)}{2 \mathrm{n}}
\end{array}\)
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