Answer:
a) n2 = 1.55
b) 408.25 nm
c) 4.74*10^14 Hz
d) 1.93*10^8 m/s
Explanation:
a) To find the index of refraction of the syrup solution you use the Snell's law:
[tex]n_1sin\theta_1=n_2sin\theta_2[/tex] (1)
n1: index of refraction of air
n2: index of syrup solution
angle1: incidence angle
angle2: refraction angle
You replace the values of the parameter in (1) and calculate n2:
[tex]n_2=\frac{n_1sin\theta_1}{sin\theta_2}=\frac{(1)(sin30.2\°)}{sin18.82\°}=1.55[/tex]
b) To fond the wavelength in the solution you use:
[tex]\frac{\lambda_2}{\lambda_1}=\frac{n_1}{n_2}\\\\\lambda_2=\lambda_1\frac{n_1}{n_2}=(632.8nm)\frac{1.00}{1.55}=408.25nm[/tex]
c) The frequency of the wave in the solution is:
[tex]v=\lambda_2 f_2\\\\f_2=\frac{v}{\lambda_2}=\frac{c}{n_2\lambda_2}=\frac{3*10^8m/s}{(1.55)(408.25*10^{-9}m)}=4.74*10^{14}\ Hz[/tex]
d) The speed in the solution is given by:
[tex]v=\frac{c}{n_2}=\frac{3*10^8m/s}{1.55}=1.93*10^8m/s[/tex]
