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The critical angle for total internal re

ection at a liquid-air interface is
42.5.
(a) If a ray of light traveling in the liquid has an angle of incidence at the
interface of 35.0, what angle does the refracted ray in the air make
with the normal?
(b) If a ray of light traveling in air has an angle of incidence at the
interface of 35.0, what angle does the refracted ray in the liquid make
with the normal?

User FrankyBoy
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1 Answer

4 votes
Answer:

(a) If the critical angle for total internal reflection at a liquid-air interface is 42.5, and a ray of light traveling in the liquid has an angle of incidence at the interface of 35.0, then the angle of refraction in the air can be found using Snell's law:

n₁ sin θ₁ = n₂ sin θ₂

where n₁ is the refractive index of the liquid and n₂ is the refractive index of air, which is approximately 1.00.

Rearranging this equation gives:

sin θ₂ = (n₁/n₂) sin θ₁

Plugging in the values gives:

sin θ₂ = (n₁/1.00) sin 35.0
sin θ₂ = (n₁/1.00) (0.5736)
sin θ₂ = 0.5736n₁

To find θ₂, we need to take the inverse sine of both sides:

θ₂ = sin⁻¹(0.5736n₁)

Substituting n₁ = 1.33 (refractive index of water) gives:

θ₂ = sin⁻¹(0.5736 × 1.33)
θ₂ = 46.5 degrees

Therefore, the refracted ray in the air makes an angle of 46.5 degrees with the normal.

(b) If a ray of light traveling in air has an angle of incidence at the interface of 35.0, the angle of refraction in the liquid can be found using the same formula and approach as above:

n₁ sin θ₁ = n₂ sin θ₂

where n₁ is the refractive index of air, which is approximately 1.00, and n₂ is the refractive index of the liquid, which is 1.33.

sin θ₂ = (n₁/n₂) sin θ₁
sin θ₂ = (1.00/1.33) sin 35.0
sin θ₂ = (0.7519)(0.5736)
sin θ₂ = 0.431

θ₂ = sin⁻¹(0.431)
θ₂ = 25.8 degrees

Therefore, the refracted ray in the liquid makes an angle of 25.8 degrees with the normal.
User Knuton
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