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

The radius is
`r = 3.1905 \ m`

Step-by-step 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) [(1)/(n_i) ]`

substituting values

`\theta = sin ^(-1) [(1)/(131) ]`

`\theta = 49.76^o`

Using SOHCAHTOA rule we have that

`tan \theta = ( r)/(d)`

=>
`r = d * tan \theta`

substituting values

`r = 2.7 * tan (49.76)`

`r = 3.1905 \ m`