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: Use the path difference:
Bright fringe: path difference $= m\lambda_1$. Dark fringe: path difference $= \left(n - \frac12\right)\lambda_2$.

Step 2: At the same point the path difference is the same:
$m\lambda_1 = \left(n - \frac12\right)\lambda_2 = \frac{(2n-1)}{2}\lambda_2$.

Step 3: Rearrange:
$\frac{\lambda_2}{\lambda_1} = \frac{2m}{2n - 1}$.

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