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

Answer: 12

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 $S=\frac{(\mu -1)tD}{d}...\left(ii\right)$
Given: $S=10\beta ...\left(iii\right)$
From equations (i), (ii) and (iii), we get
$\frac{(\mu -1)tD}{d}=10\frac{\lambda D}{d}$
$\Rightarrow \mu -1=\frac{10\lambda }{t}=\frac{10\times 6000\times 10^{-8}cm}{3\times 10^{-3}cm}$
$\Rightarrow \mu -1=0.2$
$\Rightarrow \mu =1.2=\frac{12}{10}$