Question

Interference fringes were produced in Youngs double slit experiment using light of wavelength $5000 \mathring{A}$. When a film of material $2.5 \times 10^{-3}\, cm$ thick was placed over one of the slits, the fringe pattern shifted by a distance equal to $20$ fringe widths. The refractive index of the material of the film is

• A. $1.25$
• B. $1.33$
• C. $1.4$
• D. $1.5$

Answer 1

Answer: (C)

Fringe width, $\beta = \frac{\lambda D}{d} \quad ...(i)$
where $D$ is the distance between the screen and slit and $d$ is the distance between two slits.
When a film of thickness $t$ and refractive index $\mu$ is placed over one of the slit, the fringe pattern is shifted by distance $S$ and is given by
$S = \frac{(\mu - 1) t D}{d} \quad ...(i)$
Given : $S = 20 \beta \quad ...(iii)$
From equations $(i), (ii)$ and $(iii)$ we get,
$(\mu -1) t = 20 \lambda $
or $(\mu -1) = \frac{20\lambda}{t}$
$= \frac{20\times 5000 \times 10^{-8}\,cm}{2.5 \times 10^{-3}\,cm}$
$\mu - 1 = 0.4 $ or $\mu = 1.4$