The critical angle for light wave going from a medium in which wavelength is $4000 A^o$to a medium in which its wavelength is $6000 A^o$is

Given that light is travelling from denser medium ${\mathrm{\mu }}_1$ to rarer medium ${\mathrm{\mu }}_2$ then ${\mu }_1 sin C={\mu }_2 sin 90$ (at critical angle C)

${\mu }_1=\frac{c}{v_1}=\frac{c}{n{\lambda }_1}, {\mu }_2=\frac{c}{v_2}=\frac{c}{n{\lambda }_2}$(here c is velocity of light)

So we get$\frac{c}{n{\lambda }_1} \sin C=\frac{c}{n{\lambda }_2}\Rightarrow \frac{1}{\sin C}=\frac{{\lambda }_1}{{\lambda }_2}$             

$\Rightarrow {\mathrm{\lambda }}_1=4000 {\mathrm{A}}^{\mathrm{o}}(\mathrm{denser} \mathrm{medium})$

$\Rightarrow {\mathrm{\lambda }}_2=6000 {\mathrm{A}}^{\mathrm{o}}(\mathrm{rarer} \mathrm{medium})$

$\frac{1}{\sin \mathrm{C}}=\frac{4000}{6000}\Rightarrow \sin \mathrm{C}=\frac{2}{3}\Rightarrow \mathrm{C} = {\sin }^{-1}\left(\frac{2}{3}\right)$