Answer:
a
[tex]\theta _2 = 13^o[/tex]
b
[tex]\theta _1 =32.94^o[/tex]
c
[tex]\theta_c = 53.05^o[/tex]
Explanation:
From the question we are told that
The angle of incidence is [tex]\theta_1 = 10^o[/tex]
The refractive index of water is [tex]n_1 = 1.3[/tex]
Generally Snell's law is mathematically represented as
[tex]n_1 sin(\theta_1) = n_2 sin(\theta_ 2)[/tex]
Here [tex]n_2[/tex] is the refractive index of air with value [tex]n_2 = 1[/tex]
[tex]\theta_2[/tex] is the angle of refraction
So
[tex]\theta _2 = sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ][/tex]
=> [tex]\theta _2 = sin^{-1}[\frac{1.3 * sin(10)}{1} ][/tex]
=> [tex]\theta _2 = 13^o[/tex]
Given that the angle should not be greater than [tex]\theta _2 =45^o[/tex] then the angle of incidence will be
[tex]\theta _1 = sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ][/tex]
=> [tex]\theta _1 = sin^{-1}[\frac{1 * sin(45)}{1.3} ][/tex]
=> [tex]\theta _1 =32.94^o[/tex]
Generally for critical angle is mathematically represented as
[tex]\theta_c = sin^{-1}[\frac{n_2}{n_1} ][/tex]
=> [tex]\theta_c = sin^{-1}[\frac{1}{1.3} ][/tex]
=> [tex]\theta_c = 53.05^o[/tex]
