Agreen light is submerged 2.90 m beneath the surface of a liquid with an index of refraction 1.35. what is the radius of the circle from which light escapes from the liquid into the air above the surface?

radius is 3.197 m

Explanation:

Given data

index of refraction  =  1.35

submerged  h = 2.90 m

to find out

radius of the circle

solution

we know here that light will be escape only if angle of critical is less than angle of incidence

so we can say

sin (θ)i = 1 / index of refraction

here angle (θ)i is critical angle so

sin (θ)i = 1 / 1.35 = 0.7407

(θ)i = 47.79 degree

we consider here radius r of circle

so

tan (θ)i = R /h

so

R = 2.90 tan ( 47.79)

R = 3.197 m

so radius is 3.197 m

if the thermos has a vacuum innner, then heat through conv currents and conduction are reduced = vacuum flask

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