A Young's double slit experiment is performed in air using a...

The fringe width in air is $\beta = \dfrac{\lambda D}{d}$.

The length of the field of view containing $n$ fringes is $y = n\beta = n \dfrac{\lambda D}{d}$.

When the apparatus is immersed in a liquid of refractive index $\mu$, the wavelength of light in the liquid becomes $\lambda' = \dfrac{\lambda}{\mu}$.

Since the field of view $y$ remains the same, the new number of fringes $n'$ is given by $y = n' \dfrac{\lambda' D}{d}$.

Equating the two expressions for $y$ gives $n \lambda = n' \lambda'$.

Substituting $\lambda' = \dfrac{\lambda}{\mu}$, we get $n \lambda = n' \left(\dfrac{\lambda}{\mu}\right)$, which simplifies to $n' = n \mu$.

Given $n = 12$ and $\mu = \dfrac{4}{3}$, the new number of fringes is $n' = 12 \times \dfrac{4}{3} = 16$.

Answer: $16$