Question

When a plastic thin film of refractive index 1.45 is placed in the path of one of the interfering waves then the central fringe is displaced through width of five fringes. The thickness of the film, if the wavelength of light is \( 5890 \, \text{Å} \), will be

• A. \( 6.544 \times 10^{-4} \, \text{cm} \)
• B. \( 5.644 \times 10^{-4} \, \text{m} \)
• C. \( 6.54 \times 10^{-4} \, \text{cm} \)
• D. \( 6.5 \times 10^{-4} \, \text{cm} \)

Hint:

The displacement in interference fringes due to a thin film is proportional to its thickness and the refractive index.

Answer 1

The Correct Option is A

Step 1: Fringe Displacement.
The displacement of the central fringe is related to the thickness \( t \) of the film by the equation: \[ \Delta x = \frac{2 t n}{\lambda} \] where \( n \) is the refractive index of the film, and \( \lambda \) is the wavelength of light. Substituting the given values:
Step 2: Conclusion.
The correct answer is (A), \( 6.544 \times 10^{-4} \, \text{cm} \).