Answer
$\frac{\lambda_{ethyl,alcohol}}{\lambda_{carbon,disulfide}}=\frac{1.632}{1.362}=1.198$
Work Step by Step
From table 26.1.
Refractive index for ethyl alcohol $n_{ethyl,alcohol}=1.362$
Refractive index for carbon disulfide $n_{carbon,disulfide}=1.632$
Given that fequency of light in both ethyl alcohol and carbon disulfide is $f$
refractive index is defined as
$n=\frac{c}{v}$
$c$=speed of light in vacuum
$v$= speed of light in material
we can rewrite above equation as
$v=\frac{c}{n}$
speed of light $v=f\lambda$
so $f\lambda=\frac{c}{n}$ or $\lambda=\frac{c}{nf}$.......equation(1)
suppose for ethyl alcohol wavelength is $\lambda_{ethyl,alcohol}$
so from equation(1) $\lambda_{ethyl,alcohol}=\frac{c}{n_{ethyl,alcohol}f}$........equation(2)
suppose for carbon disulfide wavelength is $\lambda_{carbon,disulfide}$
so from equation(1) $\lambda_{carbon,disulfide}=\frac{c}{n_{carbon,disulfide}f}$........equation(3)
dividing equation (2) by equation(3) will give us
$\frac{\lambda_{ethyl,alcohol}}{\lambda_{carbon,disulfide}}=
\frac{\frac{c}{n_{ethyl,alcohol}f}}{\frac{c}{n_{carbon,disulfide}f}}$
so
$\frac{\lambda_{ethyl,alcohol}}{\lambda_{carbon,disulfide}}=\frac{n_{carbon,disulfide}}{n_{ethyl,alcohol}}$
$\frac{\lambda_{ethyl,alcohol}}{\lambda_{carbon,disulfide}}=\frac{1.632}{1.362}=1.198$
