YDSE is conducted using light of wavelength 6000 ∘ A to observe an...

Fringe width, $\beta = \dfrac{\lambdaD}{d} . . . \left(i\right)$
where, D: distance between screen and slit
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 $S = \dfrac{( \mu - 1 ) tD}{d} . . . \left(ii\right)$

Given: $S = 10\beta . . . \left(iii\right)$
From equations (i), (ii) and (iii), we get

$\dfrac{( \mu - 1 ) + D}{d} = 10\dfrac{\lambdaD}{d}$

$⇒ \mu - 1 = \dfrac{10\lambda}{t} = \dfrac{10 \times 6000 \times {10}^{- 9}cm}{3 \times {10}^{- 3}cm}$

$⇒ \mu - 1 = 0 . 2$

$⇒ \mu = 1 . 2$