Question

A green light is submerged 2.70 m beneath the surface of a liquid with an index of refraction 1.31. What is the radius of the circle from which light escapes from the liquid into the air above the surface

Answer 1

Answer:

The radius is  $r = 3.1905 \ m$

Explanation:

From the question we are told that

        The  distance  beneath the liquid  is  $d = 2.70 \ m$

        The refractive index of the liquid is  $n_i = 1.31$

Now the critical value is mathematically represented as

         $\theta = sin ^{-1} [\frac{1}{n_i} ]$

substituting values

         $\theta = sin ^{-1} [\frac{1}{131} ]$

         $\theta = 49.76^o$

Using SOHCAHTOA rule we have that

         $tan \theta = \frac{ r}{d}$

=>     $r = d * tan \theta$

substituting values  

        $r = 2.7 * tan (49.76)$

        $r = 3.1905 \ m$