Let $\lambda_a$ be the wavelength of light in air and $\lambda_m$ be its wavelength in the given medium. Position of the $8^{\text{th}}$ bright fringe in the medium is given by: $y_8 = 8\beta_m = \frac{8\lambda_m D}{d}$ Position of the $5^{\text{th}}$ dark fringe in air is given by: $y_5' = (2(5)-1) \frac{\lambda_a D}{2d} = \frac{9\lambda_a D}{2d} = 4.5 \frac{\lambda_a D}{d}$ According to the question, $y_8 = y_5'$ $\frac{8\lambda_m D}{d} = 4.5 \frac{\lambda_a D}{d}$ $8\lambda_m = 4.5\lambda_a$ $\frac{\lambda_a}{\lambda_m} = \frac{8}{4.5} = \frac{16}{9}$ The refractive index of the medium is given by $\mu = \frac{\lambda_a}{\lambda_m}$. $\mu = \frac{16}{9} \approx 1.777... \approx 1.78$.