Question

In Young's double slit experiment, two slits \(S_1\) and \(S_2\) are 'd' distance apart and the separation from slits to screen is \(D\). Now if two transparent slabs of equal thickness \(0.1\) mm but refractive index \(1.51\) and \(1.55\) are introduced in the path of beam (\(λ = 4000\) Å) from \(S_1\) and \(S_2\) respectively. The central bright fringe spot will shift by ___ number of fringes.

• A. \(5\)
• B. \(10\)
• C. \(15\)
• D. \(20\)

Hint:

Shift in fringes \(=\frac{(\mu_2-\mu_1)t}{\lambda}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A slab of thickness \(t\) and index \(\mu\) adds an extra optical path \((\mu-1)t\). With slabs on both slits, the net extra path difference is \((\mu_2 - \mu_1)t\).

Step 2: Path difference:
\(\Delta x = (1.55 - 1.51)\times0.1\ \text{mm} = 0.04\times10^{-4}\ \text{m} = 4\times10^{-6}\) m.

Step 3: Number of fringes:
\(\lambda = 4000\) \(\text{\AA} = 4\times10^{-7}\) m.
\[ n = \frac{\Delta x}{\lambda} = \frac{4\times10^{-6}}{4\times10^{-7}} = 10 \]

Final Answer:
The central fringe shifts by \(10\) fringes, option (B). \[ \boxed{10} \]