Question

7. A ray of light is incident obliquely on
medium 1 and then passes through two
media namely 2 and 3 before emerging into
medium 1 again.

60%

45°

2

30d

3

The refractive index of medium 3 with
respect to medium 1 is:​

Answer 1

Answer:

n21 = sin i / sin r = sin60°/ sin45°

= √3/2 / 1/√2

= √3/√2

similarly

n32 = sin45°/sin 30°

= 1/√2 / 1/2

= √2

n31 = n32 x n21

= √3/√2 x √2

= √3

Answer 2

A ray of light is incident obliquely on medium 1 and then passes through two media namely 2 and 3 before emerging into medium 1 again.

We have to find the refractive index of medium 3 with respect to 1.

From Snell's law,

$\mu_1sin\theta_i=\mu_2sin\theta_r$

μ₁ = refractive index of medium of incident light

μ₂ = refractive index of medium of refracted light

θ_i = angle of incident light

θ_r = angle of refracted light

case 1 : when a ray of light is moving from medium 1 to medium 2.

⇒μ₁sin60° = μ₂sin45°

⇒μ₁/μ₂ = sin45°/sin60° = √(2/√3) ...(1)

case 2 : when the ray of light is moving from medium 2 to medium 3.

⇒μ₂sin45° = μ₃sin30°

⇒μ₂/μ₃ = sin30°/sin45° = 1/√2 ...(2)

from equations (1) and (2) we get,

μ₁/μ₂ × μ₂/μ₃ = μ₁/μ₃ = √(2/√3) × 1/√2 = √3

⇒μ₃/μ₁ = 1/√3

Therefore the refractive index of medium 3 with respect to medium 1 is 1/√3.