Answer
This substance is Carbon disulfide with $n=1.632$.
Work Step by Step
Given wavelength of the light in substance
$\lambda= 340nm=340\times10^{-9}m=3.4\times10^{-7}m$
Given frequency of light in substance
$f=5.403\times10^{14}Hz=5.403\times10^{14}/s$
speed of light in substance $v=f\lambda$
so $v=5.403\times10^{14}/s\times3.4\times10^{-7}m=1.837\times10^{8}m/s$
refractive index is defined as
$n=\frac{c}{v}$
$c$=speed of light in vacuum
$v$= speed of light in substance
speed of light in vacuum $c=3\times10^8m/s$
speed of light in substance $v=1.837\times10^{8}m/s$
refractive index of the substance $n=\frac{3\times10^8m/s}{1.837\times10^{8}m/s}=1.633$
from table 26.1 Carbon disulfide is having $n=1.632$ ,so this substance is Carbon disulfide.
