A ray of light passes from air into water. If the angle of incidence is 30 degrees, what is the angle of refraction in water? (Refractive index of water = 1.33)
Practice Questions
1 question
Q1
A ray of light passes from air into water. If the angle of incidence is 30 degrees, what is the angle of refraction in water? (Refractive index of water = 1.33)
22.1 degrees
30 degrees
36.9 degrees
45 degrees
Using Snell's law, n1*sin(θ1) = n2*sin(θ2), we find θ2 = 22.1 degrees.
Questions & Step-by-step Solutions
1 item
Q
Q: A ray of light passes from air into water. If the angle of incidence is 30 degrees, what is the angle of refraction in water? (Refractive index of water = 1.33)
Solution: Using Snell's law, n1*sin(θ1) = n2*sin(θ2), we find θ2 = 22.1 degrees.
Steps: 9
Step 1: Identify the refractive indices. The refractive index of air (n1) is approximately 1.0, and the refractive index of water (n2) is 1.33.
Step 2: Identify the angle of incidence (θ1), which is given as 30 degrees.
Step 3: Write down Snell's law formula: n1 * sin(θ1) = n2 * sin(θ2).
Step 4: Substitute the known values into the formula: 1.0 * sin(30 degrees) = 1.33 * sin(θ2).
Step 5: Calculate sin(30 degrees), which is 0.5. So the equation becomes: 1.0 * 0.5 = 1.33 * sin(θ2).
Step 6: Simplify the equation: 0.5 = 1.33 * sin(θ2).
Step 7: Solve for sin(θ2) by dividing both sides by 1.33: sin(θ2) = 0.5 / 1.33.
Step 8: Calculate sin(θ2): sin(θ2) ≈ 0.3759.
Step 9: Find θ2 by taking the inverse sine (arcsin) of 0.3759: θ2 ≈ 22.1 degrees.
