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 $D$ (as shown in figure) 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 $(\lambda=4000 \mathring {A})$ from $S_1$ and $S_2$ respectively, The central bright fringe spot will shift by ______ number of fringes

Answer 1

Correct Answer: 10

To determine the shift in the central bright fringe in Young's double slit experiment due to the introduction of the slabs, we calculate the optical path difference introduced by each slab. The optical path difference $\Delta x$ is given by:
\(\Delta x = (n_2-n_1)t\)
Here, \(n_1=1.51\) and \(n_2=1.55\), and the thickness \(t=0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m}\).
So, \(\Delta x = (1.55-1.51) \times 0.1 \times 10^{-3}\)
\(\Delta x = 0.04 \times 0.1 \times 10^{-3} = 4 \times 10^{-6} \, \text{m}\)
The number of fringes shifted (\(N\)) is given by \(\frac{\Delta x}{\lambda}\):
\(\lambda = 4000 \, \mathring {A} = 4000 \times 10^{-10} \, \text{m}\)
\(N = \frac{4 \times 10^{-6}}{4000 \times 10^{-10}} = \frac{4 \times 10^{-6}}{4 \times 10^{-7}} = 10\)
The central bright fringe shifts by 10 fringes, which matches the expected range of 10,10, confirming the accuracy.