Question

A double slit experiment is performed with light of wavelength \(500\) nm. A thin film of thickness \(2\) \(\mu\)m and refractive index \(1.5\) is introduced in the path of the upper beam. The location of the central maximum

• A. remain unshifted
• B. shift downward by nearly two fringes
• C. shift upward by nearly two fringes
• D. shift downward by 10 fringes

Hint:

The film adds an extra optical path of (mu - 1)t; dividing by the wavelength gives the number of fringes shifted toward the film side.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
A thin film of thickness \(t\) and refractive index \(\mu\) in one beam adds an extra optical path \((\mu-1)t\).
The whole fringe pattern shifts toward the side of the film.

Step 2: Number of fringes shifted
Number of fringes \(=\dfrac{(\mu-1)t}{\lambda}\)

Step 3: Calculation
\(\dfrac{(1.5-1)\times 2\times 10^{-6}}{500\times 10^{-9}}=\dfrac{10^{-6}}{5\times 10^{-7}}=2\)

Step 4: Compare the options
The film is on the upper beam, so the central maximum shifts upward by about two fringes, which is option (C).

Final Answer:
The central maximum shifts upward by nearly two fringes. \[ \boxed{\text{(C) }\text{shift upward by nearly two fringes}} \]