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(A) Let the distance be $s$. In medium $A$,the wavelength is $\lambda_{1}$ and in medium $B$,the wavelength is $\lambda_{2}$.Since $x$ waves are in di

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$\frac{x}{y}$

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(A) Let the distance be $s$. In medium $A$,the wavelength is $\lambda_{1}$ and in medium $B$,the wavelength is $\lambda_{2}$.Since $x$ waves are in distance $s$ in medium $A$,we have $s = x \lambda_{1}$.Since $y$ waves are in distance $s$ in medium $B$,we have $s = y \lambda_{2}$.Equating the two expressions for $s$,we get $x \lambda_{1} = y \lambda_{2}$,which implies $\frac{\lambda_{1}}{\lambda_{2}} = \frac{y}{x}$.We know that the refractive index $n$ is inversely proportional to the wavelength $\lambda$ (i.e.,$n \propto \frac{1}{\lambda}$).Therefore,the refractive index of medium $A$ with respect to medium $B$ is given by $n_{AB} = \frac{n_{A}}{n_{B}} = \frac{\lambda_{2}}{\lambda_{1}}$.Substituting the ratio of wavelengths,we get $n_{AB} = \frac{x}{y}$.

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