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

Answer: 6

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, $\theta =\frac{D\lambda }{d\left(D\right)}=\frac{\lambda }{d}$
$\frac{{\left(\theta \right)}_1}{{\left(\theta \right)}_2}=\left(\frac{{\left(\lambda \right)}_1}{{\left(\lambda \right)}_2}\right)\left(\frac{d_2}{d_1}\right)$
$\frac{0.04}{{\left(\theta \right)}_2}=\frac{\lambda }{\left(\frac{\lambda }{\mu }\right)}\left(\frac{\left(\frac{d}{2}\right)}{d}\right)\Rightarrow \theta =0.06$