What is the critical angle of light traveling from benzene (n=1.501) into air?
Choose one answer.
a. 40.05o
b. 48.22o
c. 41.78o
d. 62.63o
What is the formula for this? This is one of the simplest physics formulas there is.
Beauty
To find the critical angle of light traveling from benzene into air, we can use Snell's law. Snell's law relates the angles of incidence and refraction to the refractive indices of the two materials.
The formula for Snell's law is: n₁sinθ₁ = n₂sinθ₂
In this case, n₁ is the refractive index of benzene (1.501), n₂ is the refractive index of air (approximately 1.000), θ₁ is the angle of incidence (which we are trying to find), and θ₂ is the angle of refraction (which would be 90° at the critical angle).
At the critical angle, the angle of refraction is 90°, so sinθ₂ = 1.
Plugging these values into Snell's law, we get: 1.501sinθ₁ = 1sin90° = 1.
Now we can solve for sinθ₁: sinθ₁ = 1/1.501 ≈ 0.666.
To find the angle θ₁, we can take the inverse sine (sin⁻¹) of 0.666: θ₁ ≈ sin⁻¹(0.666) ≈ 41.78°.
Therefore, the correct answer is c. 41.78°.