Critical Angle for light moving from medium I to medium 2 is θ. The s

1.	Critical Angle for light moving from medium I to medium 2 is θ. The speed of light in medium II is v then the speed in medium 2 is A) v (1-cosθ) B) v/sinθ C) v/cosθ D) v (1- sinθ)
Answer» The correct option is B) v/sinθ. From Snell's Law: n_₁sini = n_₂ sinr..................... (i) Where n₁ =Refractive index of first medium Refractive index of second medium I = angle of incidence r = angle of refraction For critical angle: Angle of incidence = i = θ Angle of refraction = r = 90° By substituting in equation 1 we get: n_₁sinθ = n_₂sin90° n_₁sin θ = n_₂ x 1 Sinθ = n_₂/n_₁.................... (ii) But as we already know the equation= n = c/V Where n inversely proportional to V If v and v₂ are the speeds of first medium and second medium then : n_₂/n_₁= v/v₂.............. (iii) By equation ii and iii we get, Sinθ = v/v₂ v₂sinθ = V ⇒ v₂ = V/sinθ

Critical Angle for light moving from medium I to medium 2 is θ. The speed of light in medium II is v then the speed in medium 2 is A) v (1-cosθ) B) v/sinθ C) v/cosθ D) v (1- sinθ)

Answer»

The correct option is B) v/sinθ.

From Snell's Law:

n_₁sini = n_₂ sinr..................... (i)

Where n₁ =Refractive index of first medium

Refractive index of second medium

I = angle of incidence

r = angle of refraction

For critical angle:

Angle of incidence = i = θ

Angle of refraction = r = 90°

By substituting in equation 1 we get:

n_₁sinθ = n_₂sin90°

n_₁sin θ = n_₂ x 1 Sinθ = n_₂/n_₁.................... (ii)

But as we already know the equation= n = c/V

Where n inversely proportional to V

If v and v₂ are the speeds of first medium and second medium then :

n_₂/n_₁= v/v₂.............. (iii)

By equation ii and iii we get,

Sinθ = v/v₂

v₂sinθ = V

⇒ v₂ = V/sinθ