Question

In a biprism experiment, fifth dark fringe is obtained at a point. A thin transparent film of refractive index '\(\mu\)' is placed in one of the interfering paths. Now \(7^{\text{th}}\) bright fringe is obtained at the same point. If '\(\lambda\)' is the wavelength of light used, the thickness of film is equal to

• A. \(1.5(\mu - 1)\lambda\)
• B. \(\frac{1.5\lambda}{(\mu - 1)}\)
• C. \(2.5(\mu - 1)\lambda\)
• D. \(\frac{2.5\lambda}{(\mu - 1)}\)

Hint:

Film shift: $(\mu -1)t$ changes path difference

Answer 1

The Correct Option is D

Concept:
• Dark fringe: $\delta = \frac{(2n-1)\lambda}{2}$
• Bright fringe: $\delta = n\lambda$
• Film introduces shift: $(\mu -1)t$

Step 1: Initial path difference. \[ \delta = \frac{(2\times5 -1)\lambda}{2} = \frac{9\lambda}{2} \]

Step 2: After film insertion. \[ \delta' = 7\lambda \]

Step 3: Extra path difference. \[ (\mu -1)t = \delta' - \delta = 7\lambda - \frac{9\lambda}{2} = \frac{5\lambda}{2} \]

Step 4: Solve. \[ t = \frac{2.5\lambda}{\mu -1} \] Final Answer: Option (D)