Question

YDSE is conducted using light of wavelength $6000\,\mathring{A}$ to observe interference pattern. When a film of material $3.0\times 10^{- 3}cm$ thick was placed over one of the slits, the fringe pattern shifted by a distance equal to $10$ fringe widths. If the refractive index of the material of the film is $\frac{x}{10}$ then find $x$ .

Answer 1

Fringe width, $\beta =\frac{\lambda D}{d}...\left(i\right)$
where, D: distance between screen and slits
d: 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 shift by distance S and is given by $S=\frac{\left(\right. \mu - 1 \left.\right) tD}{d}...\left(i i\right)$
Given: $S=10\beta ...\left(i i i\right)$
From equations (i), (ii) and (iii), we get
$\frac{\left(\right. \mu - 1 \left.\right) t D}{d}=10\frac{\lambda D}{d}$
$\Rightarrow \mu -1=\frac{10 \lambda }{t}=\frac{10 \times 6000 \times 10^{- 8} cm}{3 \times 10^{- 3} cm}$
$\Rightarrow \mu -1=0.2$
$\Rightarrow \mu =1.2=\frac{12}{10}$