Rohan
Last Activity: 6 Years ago
A sketch will show you the vector (-1)î + (-2)ĵ makes an angle θ₁ to the normal of the interface where
tan(θ₁) = (-1)/(-2) = 1/2
θ₁ = tan⁻¹(1/2)
If you sketch a right angled triangle with perpendicular sides of lengths 1 and 2 you will see the hypotenuse is length √(1² +2²) = √5. It follows that sin(θ₁) = 1/√5
If the angle of refraction is θ₂ we can apply Snell's Law:
n₁sin(θ₁) =n₂sin(θ₂)
2 X 1/√5 =√5/2 x sin(θ₂)
2 = 5/2 x sin(θ₂)
θ₂ = sin⁻¹(4/5).
We recognise that in a '3, 4, 5 triangle θ₂ = sin⁻¹(4/5),.means that tan(θ₂) = 4/3.
From our initial refraction diagram, this means the refracted ray must lie along the vector (-4)î + (-3)ĵ.
Since √(4² +3²) = 5, the unit vector along which the refracted light ray moves is (1/5)[(-4)î + (-
