In the Figure below, Ray 1 is the incident ray with the angle 01, measured from the normal (dotted line). The incident ray moves in medium 1 with n1. Use the figure and table below for owing questions. rays 1 3 n1 n2 water 1.333 glass (flint) 1.66 diamond 2.419 ethanol 1.361 ice 1.309 If the incident ray 0, = 15° and it moves from ice to air, which of these follows for the refracted ray 02? O 02 > 15° O 02 < 15° O 02 = 15° Cannot be determined.
Refraction of Light
Refraction is a change in the direction of light rays when they travel from one medium to another. It is the bending of light when it goes through different media.
Angle of Refraction
Light is considered by many scientists to have dual nature, both particle nature and wave nature. First, Particle nature is one in which we consider a stream of packets of energy called photons. Second, Wave nature is considering light as electromagnetic radiation whereas part of it is perceived by humans. Visible spectrum defined by humans lies in a range of 400 to 700 nm wavelengths.
Index of Refraction of Diamond
Diamond, the world’s hardest naturally occurring material and mineral known, is a solid form of the element carbon. The atoms are arranged in a crystal structure called diamond cubic. They exist in a huge variety of colours. Also, they are one of the best conductors of heat and have a very high melting point.


Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 3 images









