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: Plan:
Use the fringe width and shift in position.

Step 2: Steps:
Central fringe shift $= \frac{D}{d}(\mu_2-\mu_1)t$, and fringe width $\beta = \frac{\lambda D}{d}$. So the number of fringes is $\frac{(\mu_2 - \mu_1)t}{\lambda}$.
$= \frac{0.04\times10^{-4}}{4\times10^{-7}} = 10$ fringes.

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