Answer with Explanation:
We are given that
Angle of incidence,[tex]i=30^{\circ}[/tex]
Angle of refraction,[tex]r=19.24^{\circ}[/tex]
a.Refractive index of air,[tex]n_1=1[/tex]
We know that
[tex]n_2sinr=n_1sini[/tex]
[tex]n_2=\frac{n_1sin i}{sin r}=\frac{sin30}{sin19.24}=1.517[/tex]
b.Wavelength of red light in vacuum,[tex]\lambda=632.8nm=632.8\times 10^{-9} m[/tex]
[tex]1nm=10^{-9} m[/tex]
Wavelength in the solution,[tex]\lambda'=\frac{\lambda}{n_2}[/tex]
[tex]\lambda'=\frac{632.8}{1.517}=417nm[/tex]
c.Frequency does not change .It remains same in vacuum and solution.
Frequency,[tex]\nu=\frac{c}{\lamda}=\frac{3\times 10^8}{632.8\times 10^{-9}}[/tex]
Where [tex]c=3\times 10^8 m/s[/tex]
Frequency,[tex]\nu=4.74\times 10^{14}Hz[/tex]
d.Speed in the solution,[tex]v=\frac{c}{n_2}[/tex]
[tex]v=\frac{3\times 10^8}{1.517}=1.98\times 10^8m/s[/tex]
