Water and glass with refractive index of 1.5 are interface. If the incident ray from the water makes an angle of 60 degree with the normal find the direction of the reflected and refracted ray
Take refractive index of water to 1.33
reflection same angle as incident
refraction follows Snell's Law.
To find the direction of the reflected and refracted rays at the interface between water and glass, we can use the laws of reflection and refraction.
1. Law of Reflection:
According to the law of reflection, the angle of incidence is equal to the angle of reflection, measured with respect to the normal.
Given:
Angle of incidence (θi) = 60°
Therefore, the angle of reflection (θr) will also be 60°.
2. Snell's Law of Refraction:
According to Snell's law, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the refractive indices of the two media.
Given:
Refractive index of water (n1) = 1.33
Refractive index of glass (n2) = 1.5
Angle of incidence (θi) = 60°
Using Snell's law: n1sin(θi) = n2sin(θr)
Substituting the given values:
(1.33)sin(60°) = (1.5)sin(θr)
Calculating:
0.866*1.33 = 1.5sin(θr)
1.15478 = 1.5sin(θr)
sin(θr) = 1.15478 ÷ 1.5
sin(θr) = 0.7699
θr ≈ sin^(-1)(0.7699)
θr ≈ 50.7°
Therefore, the angle of refraction (θr) is approximately 50.7°.
Now, to determine the direction of the reflected and refracted rays:
- The reflected ray will be on the same side as the incident ray and will make an angle of 60° with the normal.
- The refracted ray will be on the opposite side of the normal as the incident ray and will make an angle of 50.7° with the normal.
To find the direction of the reflected and refracted rays at the interface between water and glass, we can use Snell's Law and the law of reflection.
Snell's Law states:
n1 * sin(theta1) = n2 * sin(theta2)
Where:
- n1 and n2 are the refractive indices of the two media (water and glass, in this case),
- theta1 is the angle of incidence (between the incident ray and the normal), and
- theta2 is the angle of refraction (between the refracted ray and the normal).
The law of reflection states that the angle of reflection (between the reflected ray and the normal) is equal to the angle of incidence.
Given:
- Refractive index of water (n1) = 1.33
- Refractive index of glass (n2) = 1.5
- Angle of incidence (theta1) = 60 degrees
First, let's find the angle of refraction (theta2) using Snell's Law:
n1 * sin(theta1) = n2 * sin(theta2)
1.33 * sin(60 degrees) = 1.5 * sin(theta2)
0.866 = 1.5 * sin(theta2)
sin(theta2) = 0.866 / 1.5
theta2 = arcsin(0.577) ≈ 35.3 degrees
Now, let's find the direction of the reflected ray:
The angle of reflection is equal to the angle of incidence, so the reflected ray will make an angle of 60 degrees with the normal, but on the opposite side.
Finally, let's find the direction of the refracted ray:
The refracted ray will bend towards the normal due to the higher refractive index of water compared to glass. The refracted ray will make an angle of approximately 35.3 degrees with the normal.
To summarize:
- The reflected ray will make an angle of 60 degrees with the normal, but on the opposite side.
- The refracted ray will make an angle of approximately 35.3 degrees with the normal, within the glass.