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(A) The refractive index is given by $\mu = c/v$,where $c$ is the speed of light in vacuum and $v$ is the speed in the medium.Given $\mu_{A}/\mu_{B} =

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$5 	imes 10^{-10} v_{A} 	ext{ m}$

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(A) The refractive index is given by $\mu = c/v$,where $c$ is the speed of light in vacuum and $v$ is the speed in the medium.Given $\mu_{A}/\mu_{B} = 1/2$,we have $(c/v_{A}) / (c/v_{B}) = v_{B}/v_{A} = 1/2$,which implies $v_{A} = 2v_{B}$.Let the thickness of both materials be $d$.The time taken to travel through the material is $t = d/v$.Given $t_{2} - t_{1} = 5 	imes 10^{-10} 	ext{ s}$,where $t_{1} = d/v_{A}$ and $t_{2} = d/v_{B}$.Substituting the values: $d/v_{B} - d/v_{A} = 5 	imes 10^{-10}$.$d(1/v_{B} - 1/v_{A}) = 5 	imes 10^{-10}$.$d((v_{A} - v_{B}) / (v_{A}v_{B})) = 5 	imes 10^{-10}$.Since $v_{A} = 2v_{B}$,we have $d((2v_{B} - v_{B}) / (2v_{B} \cdot v_{B})) = 5 	imes 10^{-10}$.$d(v_{B} / 2v_{B}^{2}) = 5 	imes 10^{-10}$.$d / (2v_{B}) = 5 	imes 10^{-10}$.$d = 10 	imes 10^{-10} v_{B} = 10^{-9} v_{B} 	ext{ m}$.Alternatively,using $v_{B} = v_{A}/2$,we get $d / (2(v_{A}/2)) = 5 	imes 10^{-10} \Rightarrow d/v_{A} = 5 	imes 10^{-10} \Rightarrow d = 5 	imes 10^{-10} v_{A} 	ext{ m}$.

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