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A beam of light starts at a point 8.0 cm beneath the surface of a liquid and strikes the surface 7.0 cm from the point directly above the li
Question
A beam of light starts at a point 8.0 cm beneath the surface of a liquid and strikes the surface 7.0 cm from the point directly above the light source. Total internal reflection occurs for the beam. What can be said about the index of refraction of the liquid?
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Physics
4 years
2021-07-26T08:56:08+00:00
2021-07-26T08:56:08+00:00 1 Answers
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Answer:
Explanation:
Given that,
The light starts at a point 8cm beneath the surface
It strikes the surface 7cm directly above the light
The angle of incidence from the given distances is
Using trigonometry
Tan I = opp / adj
tan I= (7.0 cm)/(8.0 cm) = 0.875
I = arctan(0.875)
I = 41.1859
According to Snell’s law
n(water) sin I = n(air) sin R
Here R =90° for total reflection to occur
nair = 1
n(water) sin I = n(air) sinR
n(water) = (1)sin 90
n(water) = 1
n(water) = 1/sin I
n(water) =1/ sin(41.1859o)
= 1.52
The refractive index index of the liquid is 1.52