The critical angle for total internal reflection at a liquid-air interface is 41.0. (1) if a ray of light traveling in the liquid has an angle of incidence of 34.5 at the interface with respect to the normal, what angle does the refracted ray in the air make with the normal? (2) if a ray of light traveling in air has an angle of incidence of 34.5 at the interface with respect to the normal, what angle does the refracted ray in the liquid make with the normal?

a) θ₂ = 60.9593º

b)  θ₂ = 21.8141º

Explanation:

a) Given

1 → liquid

2 → the air

θcrit = 41º

n₂ = 1

θ₁ = 35º

θ₂ = ?

We apply the formula

Sin θcrit = n₂ / n₁    ⇒    n₁ = n₂ / Sin θcrit = 1 / Sin 41º = 1.5242

Now, we apply Snell's Law

n₁*Sin θ₁ = n₂*Sin θ₂      

⇒   θ₂ = ArcSin (n₁*Sin θ₁ / n₂)

⇒   θ₂ = ArcSin (1.5242*Sin 35º / 1) = 60.9593º

b) From air to liquid

1 → the air

2 → liquid

n₁ = 1

θ₁ = 34.5º

n₂ = 1.5242

we apply Snell's Law

n₁*Sin θ₁ = n₂*Sin θ₂      

⇒   θ₂ = ArcSin (n₁*Sin θ₁ / n₂)

⇒   θ₂ = ArcSin (1*Sin 34.5º / 1.5242) = 21.8141º

more than, less

less than, more

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.