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

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°)

       θ₁ = 67.67°

But this angle is measured from the normal.

so, angle from the wall

       = 90° - 67.67°

       = 22.33°

grav potential energy