Question

The speed of light in two media ‘1’ and ‘2’ are v₁ and v₂ (> v₁) respectively. For a ray of light to undergo total internal reflection at the interface of these two media, it must be incident from ______.

Options

• medium ‘1’ and at an angle greater than sin⁻¹ `(v_1/v_2)`

• medium ‘1’ and at an angle greater than cos⁻¹ `(v_1/v_2)`

• medium ‘2’ and at an angle greater than sin⁻¹ `(v_1/v_2)`

• medium ‘2’ and at an angle greater than cos⁻¹ `(v_1/v_2)`

Answer 1

The speed of light in two media ‘1’ and ‘2’ are v₁ and v₂ (> v₁) respectively. For a ray of light to undergo total internal reflection at the interface of these two media, it must be incident from `bbunderline("medium ‘2’ and at an angle greater than sin"^(−1) (v_1/v_2))`.

Explanation:

Refractive index (μ) = `c/v`

Thus, μ₁ = `c/v_1`

μ₂ = `c/v_2`

Since, v₁ > v₂, Thus, μ₁ < μ₂

Now, for TIR (total internal reflection), light must go from a denser to a rarer medium. Thus, light must go from medium 2 to 1, and the angle of incidence must be greater than the critical angle.

Noe the critical angle (i_c) = `sin^-1(mu_1/mu_2)`

Thus, `mu_1/mu_2 = v_2/v_1`