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If a light ray enters a medium with a refractive index of 1.33 at an angle of 60
If a light ray enters a medium with a refractive index of 1.33 at an angle of 60°, what is the angle of refraction in the medium?
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If a light ray enters a medium with a refractive index of 1.33 at an angle of 60°, what is the angle of refraction in the medium?
30°
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Using Snell's law: n1 * sin(θ1) = n2 * sin(θ2), 1 * sin(60°) = 1.33 * sin(θ2) => sin(θ2) = (sin(60°)/1.33) ≈ 0.577, θ2 ≈ 35.0°.
Questions & Step-by-step Solutions
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Q
Q: If a light ray enters a medium with a refractive index of 1.33 at an angle of 60°, what is the angle of refraction in the medium?
Solution:
Using Snell's law: n1 * sin(θ1) = n2 * sin(θ2), 1 * sin(60°) = 1.33 * sin(θ2) => sin(θ2) = (sin(60°)/1.33) ≈ 0.577, θ2 ≈ 35.0°.
Steps: 10
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Step 1: Identify the refractive index of the first medium (n1). In this case, n1 = 1 (for air).
Step 2: Identify the refractive index of the second medium (n2). Here, n2 = 1.33.
Step 3: Identify the angle of incidence (θ1). In this case, θ1 = 60°.
Step 4: Use Snell's law formula: n1 * sin(θ1) = n2 * sin(θ2).
Step 5: Substitute the known values into the formula: 1 * sin(60°) = 1.33 * sin(θ2).
Step 6: Calculate sin(60°). It is approximately 0.866.
Step 7: Rewrite the equation: 0.866 = 1.33 * sin(θ2).
Step 8: Solve for sin(θ2) by dividing both sides by 1.33: sin(θ2) = 0.866 / 1.33.
Step 9: Calculate sin(θ2). This gives approximately 0.651.
Step 10: Find θ2 by taking the inverse sine (arcsin) of 0.651. This gives θ2 ≈ 40.5°.
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