Question

The glass core of an optical fiber has an index of refraction

1.60.The index of refraction of the cladding is 1.48. What is
themaximum angle a light ray can make with the wall of the core if
itis to remain inside the fiber?

Answer 1

given,

refractive index of glass (n₁)= 1.60

refractive index of cladding,(n₂) = 1.48

We need to find the critical angle θ in the core such that the refracted angle in the cladding is 90°

Using Snell's Law we have

n₁ x sin (θ₁ ) = n₂ x sin(θ₂)

θ₂ = 90°

n₁ x sin (θ₁ ) = n₂ x sin(90°)

`\theta_1 = sin^(-1)((n_2)/(n_1))`

`\theta_1 = sin^(-1)((1.48)/(1.60))`

`\theta_1 = sin^(-1)(0.925)`

θ₁ = 67.67°

But this angle is measured from the normal.

so, angle from the wall

= 90° - 67.67°

= 22.33°