Question

In a young's double slit experiment, the angular separation of interference fringes on a distant screen in $0.04$ . The angular separation if the entire apparatus is inferred in a liquid of refractive index $\frac{4}{3}$ and distance between slits become half is $0.0n\,\mathring{A}$ then what is the value of $n$ ?

Answer 1

The formula of fringe width in terms of distance between screen and slits, the separation between slits and wavelength of monochromatic light which is used, $\beta =\frac{D \lambda }{d}$ and angular width, $\theta =\frac{D \lambda }{d \left(D\right)}=\frac{\lambda }{d}$
$\frac{\left(\theta \right)_{1}}{\left(\theta \right)_{2}}=\left(\frac{\left(\lambda \right)_{1}}{\left(\lambda \right)_{2}}\right)\left(\frac{d_{2}}{d_{1}}\right)$
$\frac{0 . 04}{\left(\theta \right)_{2}}=\frac{\lambda }{\left(\frac{\lambda }{\mu }\right)}\left(\frac{\left(\frac{d}{2}\right)}{d}\right)\Rightarrow \theta =0.06$