Question

A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 49° with respect to the normal to the surface. It passes into the second medium and refracts at an angle of θ2 = 69.5° with respect to the normal.

Required:
a. Write an expression for the index of refraction of the second material.
b. Numerically, what is this index?
c. Numerically, what is the light's velocity in medium 1, in meters per second?
d. Numerically, what is the light's velocity in medium 2, in meters per second?

Answer 1

a) n2=(n1sin1)(sin2)

b) 1.18

c) 201081632.7m/s

d) 254237288.1m/s

Step-by-step explanation:

a) We can calculate the index of refraction of the second material by using the Snell's law:

`n_1sin\theta_1=n_2sin\theta_2`

`n_2=(n_1sin\theta_1)/(sin\theta_2)`

b) By replacing in the equation of a) we obtain:

`n_2=((1,47)sin49\°)/(sin69.5\°)=1.18`

c) light velocity in the medium is given by:

`v=(c)/(n_1)=(3*10^(8)m/s)/(1.47)=204081632.7(m)/(s)`

d)

`v=(c)/(n_2)=(3*10^(8)m/s)/(1.18)=254237288.1(m)/(s)`

hope this helps!!