Question

In an interference experiment, the \(m^{th}\) bright fringe for light of wavelength \(λ_1\) coincides with the \(n^{th}\) dark fringe for light of wavelength \(λ_2\) . The ratio \(\frac{λ_2}{λ_1}\) is

• A. \(\frac{m}{n-1}\)
• B. \(\frac{m}{(2n-1)}\)
• C. \(\frac{2m}{(2n-1)}\)
• D. \(\frac{2m}{(2n+1)}\)

Hint:

Bright fringe: y = m lambda D/d; dark fringe: y = (2n - 1) lambda D/(2d).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In Young's experiment, the position of the \(m^{th}\) bright fringe is \(y = \frac{m\lambda D}{d}\). The position of the \(n^{th}\) dark fringe is \(y = \frac{(2n - 1)\lambda D}{2d}\).

Step 2: Equate the positions:
\[ \frac{m\lambda_1D}{d} = \frac{(2n - 1)\lambda_2D}{2d} \]

Step 3: Solve:
\[ \frac{\lambda_2}{\lambda_1} = \frac{2m}{2n - 1} \]

Step 4: Why the other options are wrong.
\(\frac{m}{n-1}\) and \(\frac{m}{2n-1}\) lack the factor 2 from the dark fringe formula. \(\frac{2m}{2n+1}\) uses \(2n + 1\), which is the formula for the dark fringe counted from \(n = 0\), not from \(n = 1\).

Final Answer:
The ratio is \(\frac{2m}{2n - 1}\), option (C). \[ \boxed{\frac{2m}{2n-1}} \]