Answer:
The radius is [tex]r = 3.1905 \ m[/tex]
Explanation:
From the question we are told that
The distance beneath the liquid is [tex]d = 2.70 \ m[/tex]
The refractive index of the liquid is [tex]n_i = 1.31[/tex]
Now the critical value is mathematically represented as
[tex]\theta = sin ^{-1} [\frac{1}{n_i} ][/tex]
substituting values
[tex]\theta = sin ^{-1} [\frac{1}{131} ][/tex]
[tex]\theta = 49.76^o[/tex]
Using SOHCAHTOA rule we have that
[tex]tan \theta = \frac{ r}{d}[/tex]
=> [tex]r = d * tan \theta[/tex]
substituting values
[tex]r = 2.7 * tan (49.76)[/tex]
[tex]r = 3.1905 \ m[/tex]
