IGCSE Physics 0625: Critical Angle and Total Internal Reflection in an Optical Fibre

A doctor uses an endoscope to examine the inside of a patient's stomach. Light travels along a thin, flexible optical fibre inside the endoscope, repeatedly striking the boundary between the fibre's core and its surrounding cladding. At this boundary, the refractive index of the core relative to the cladding is $1.50$ .

(a) Calculate the critical angle for light travelling inside the fibre core towards the boundary with the cladding. [2]

(b) A ray of light inside the fibre strikes the core–cladding boundary at an angle of incidence of $55°$ to the normal. State and explain whether total internal reflection occurs at this point. [2]

(c) Explain why it is essential that light continues to strike the sides of the fibre at an angle greater than the critical angle along the entire length of the fibre, and describe what would happen to the light signal at any point where the fibre was bent so sharply that this condition was no longer met. [3]

(d) State one everyday application of optical fibres, other than endoscopes, that relies on total internal reflection. [1]

Part (a): Calculating the critical angle

The critical angle $c$ is related to the refractive index $n$ of the core (relative to the cladding) by:

$\sin c = \frac{1}{n}$

Substituting $n = 1.50$ :

$\sin c = \frac{1}{1.50} \approx 0.667$

$c = \sin^{-1}(0.667)$

$c \approx 41.8°$

Part (b): Checking for total internal reflection

Total internal reflection occurs whenever the angle of incidence inside the denser medium is greater than the critical angle for that boundary.

Comparing the given angle of incidence with the critical angle found in (a):

$55° > 41.8°$

Since $55°$ exceeds the critical angle, total internal reflection does occur at this point on the core–cladding boundary, the ray reflects entirely back into the core, with none of it escaping into the cladding.

Part (c): Why the angle must stay above the critical angle everywhere

An optical fibre only works because light is guided along its length by total internal reflection happening again and again every time the ray meets the side of the core. As long as each reflection keeps the angle of incidence above the critical angle, essentially all of the light’s energy stays inside the core and reaches the far end.

If the fibre were bent too sharply at some point, the direction of the fibre’s surface relative to the ray would change abruptly. This can reduce the angle of incidence at that bend to below the critical angle. At that point, the condition for total internal reflection is no longer satisfied, so instead of reflecting entirely, the ray partially or wholly refracts out through the side of the fibre into the cladding (and beyond) and is lost from the core. This weakens the light signal reaching the far end of the fibre, and a severe enough bend could lose so much light that the signal becomes unusable, which is why optical fibres must not be bent beyond their minimum safe bending radius.

Part (d): Another application of optical fibres

Optical fibres that rely on total internal reflection are also used in telecommunications, carrying telephone calls and internet data as pulses of light along fibre-optic cables over long distances.

Final answers

(a) Critical angle $\approx \boxed{41.8°}$
(b) $55° > 41.8°$ , so total internal reflection does occur
(c) Total internal reflection must occur at every point for light to stay guided along the fibre; at a sharp bend where the angle of incidence drops below the critical angle, light escapes through the side of the fibre, weakening or losing the signal
(d) Telecommunications (fibre-optic cables carrying telephone/internet signals)