Question

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

• A. \(1.5(μ-1)λ\)
• B. \(\frac{1.5\,λ}{(μ-1)}\)
• C. \(2.5(μ-1)λ\)
• D. \(\frac{2.5\,λ}{(μ-1)}\)

Hint:

Fifth dark fringe has path difference 4.5 lambda. Seventh bright fringe has 7 lambda. The film adds (mu - 1) t.

Answer 1

The Correct Option is D

Step 1: Count in half wavelengths:
Fifth dark: $9$ half-wavelengths. Seventh bright: $14$ half-wavelengths. The film added $14 - 9 = 5$ half-wavelengths of path.

Step 2: Convert to wavelengths:
$5\times\frac\lambda2 = 2.5\lambda$.

Step 3: Relate to the film:
Extra path $= (\mu-1)t = 2.5\lambda$, so $t = \dfrac{2.5\lambda}{\mu - 1}$.

Step 4: Sanity check:
A film with a larger index adds more path per unit thickness, so $t$ must fall as $\mu - 1$ grows. This agrees with the form in option (D).

Final Answer:
Option (D). \[ \boxed{\frac{2.5\lambda}{\mu-1} \text{ (D)}} \]