4. A beam of light in water enters a plastic block (index of refraction=1.40(a) if the angle of incidence is 40, what is the angle of refraction in the plastic?(b) at what angle must the beam be incident on the plastic if the refraction angle is to be 30?
To answer these questions, we can use Snell's law, which describes the relationship between the angle of incidence and the angle of refraction when light passes through the interface between two materials with different refractive indices.
The formula for Snell's law is:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
Where:
- n₁ and n₂ are the refractive indices of the initial and final mediums, respectively.
- θ₁ is the angle of incidence, and θ₂ is the angle of refraction.
(a) To find the angle of refraction in the plastic block when the angle of incidence in water is 40 degrees, we can use Snell's law.
Given:
n₁ (water) = 1.33 (approximated)
n₂ (plastic) = 1.40 (given)
θ₁ (angle of incidence) = 40 degrees
Let's solve for θ₂ (angle of refraction):
n₁ * sin(θ₁) = n₂ * sin(θ₂)
1.33 * sin(40) = 1.40 * sin(θ₂)
First, calculate the left-hand side of the equation:
1.33 * sin(40) ≈ 0.861
Now, solve for θ₂:
0.861 = 1.40 * sin(θ₂)
sin(θ₂) = 0.861 / 1.40
θ₂ ≈ arcsin(0.861 / 1.40)
θ₂ ≈ 57.60 degrees
Therefore, the angle of refraction in the plastic block is approximately 57.60 degrees when the angle of incidence in water is 40 degrees.
(b) To find the angle at which the beam must be incident on the plastic block for the refraction angle to be 30 degrees, we can rearrange Snell's law.
Given:
n₁ (water) = 1.33 (approximated)
n₂ (plastic) = 1.40 (given)
θ₂ (angle of refraction) = 30 degrees
Rearranging Snell's law, we have:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
sin(θ₁) = (n₂ / n₁) * sin(θ₂)
Substituting the given values:
sin(θ₁) = (1.40 / 1.33) * sin(30)
sin(θ₁) ≈ 1.052 * 0.500
sin(θ₁) ≈ 0.526
To solve for θ₁, we take the inverse sine (or arcsin) of 0.526:
θ₁ ≈ arcsin(0.526)
θ₁ ≈ 32.18 degrees
Therefore, the beam must be incident on the plastic block at an angle of approximately 32.18 degrees for the refraction angle to be 30 degrees.