the index of refraction of h2o is 1.33. find the angle of refraction of a beam of light in air hitting the h2o surface at an angle of 60 degrees n(air)=1.00
Snell's Law:
Angles are measured to the normal, not surface.
SinThetaincident*nair=SinThetaRefracted*nwater
sin30*1=SinTheta*1.33
theta=arcsin .377=22 deg
k so the answer would b 22 degrees????
To find the angle of refraction, we can use Snell's law, which relates the angles and indices of refraction of light as it passes from one medium to another. The formula is:
n₁sinθ₁ = n₂sinθ₂
where:
n₁ is the index of refraction of the initial medium (air)
θ₁ is the angle of incidence (the angle between the incident light ray and the normal to the surface)
n₂ is the index of refraction of the second medium (water)
θ₂ is the angle of refraction (the angle between the refracted light ray and the normal to the surface)
In this case, we are given:
n₁ (air) = 1.00
θ₁ (angle of incidence) = 60 degrees
n₂ (water) = 1.33 (the index of refraction of water)
We can substitute these values into the formula and solve for θ₂:
(1.00)sin(60) = (1.33)sin(θ₂)
To find θ₂, we can rearrange the equation:
sin(θ₂) = (1.00)sin(60) / (1.33)
Using a scientific calculator or trigonometric table, we can find the value of arcsin:
θ₂ = arcsin[((1.00)sin(60))/(1.33)]
Evaluating this expression will give us the angle of refraction θ₂ in water.