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2 public questions tagged with this topic.

In optical fibers, what ensures that light remains confined within the core during transmission?

**Spherical refracting surface** power P = (n₂-n₁)/R, lens power sum of two surfaces. Lens maker derivation combines two refractions, sign of R₂ negative for second surface convex opposite direction, yielding 1/f positive for convex lens. In optical fibers, the core has a higher refractive index than the cladding, enabling total internal reflection. When light strikes the core-cladding boundary at an angle greater than the critical angle, it reflects back into the core, ensuring confinement and minimal loss. Substituting values gives Higher refractive index of core than cladding, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f

Ref: NCERT > Physics Book > Ray Optics > Refraction at Spherical Surfaces and Lens Maker's Formula

What is the primary reason optical fibers can transmit light over long distances with minimal loss?

**Lens maker's formula** 1/f = (n-1)(1/R₁ - 1/R₂), n refractive index, R₁,R₂ radii of curvature (m), sign convention R positive if surface convex towards incident light. For double convex R₁=12 cm, R₂=-12 cm, n=1.5, 1/f=(0.5)(1/12 -1/(-12))=(0.5)(2/12)=1/12, f=12 cm, converging. Optical fibers use total internal reflection to transmit light. The core has a higher refractive index than the cladding, ensuring that light rays striking the boundary at angles greater than the critical angle are fully reflected, preventing loss of light intensity over distance. Substituting values gives Total internal reflection within the core, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v -

Ref: NCERT > Physics Book > Ray Optics > Refraction at Spherical Surfaces and Lens Maker's Formula