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(C) The number of waves $N$ in a medium of thickness $t$ and refractive index $n$ is given by $N = \frac{n t}{\lambda}$,where $\lambda$ is the wavelen

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$1.3$

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(C) The number of waves $N$ in a medium of thickness $t$ and refractive index $n$ is given by $N = \frac{n t}{\lambda}$,where $\lambda$ is the wavelength in vacuum.For slab $A$ of thickness $t$ and refractive index $n_A = 1.5$:$N_A = \frac{n_A t}{\lambda} = \frac{1.5 t}{\lambda}$ .... $(i)$For the second slab,it consists of two parts $B$ and $C$ with thicknesses $t_B = \frac{t}{3}$ and $t_C = \frac{2t}{3}$ respectively. The refractive index of $C$ is $n_C = 1.6$. Let the refractive index of $B$ be $n_B$:$N_{BC} = \frac{n_B t_B}{\lambda} + \frac{n_C t_C}{\lambda} = \frac{n_B (t/3)}{\lambda} + \frac{1.6 (2t/3)}{\lambda}$ .... $(ii)$Given that the number of waves is the same in both slabs $(N_A = N_{BC})$:$\frac{1.5 t}{\lambda} = \frac{n_B t}{3\lambda} + \frac{3.2 t}{3\lambda}$Dividing by $t/\lambda$ on both sides:$1.5 = \frac{n_B}{3} + \frac{3.2}{3}$$4.5 = n_B + 3.2$$n_B = 4.5 - 3.2 = 1.3$

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