In this experiment, you will model the behaviour of an optical fibre using a curved glass rod. You directs a ray of light from a ray box into one end of the...
Question 1 Report
In this experiment, you will model the behaviour of an optical fibre using a curved glass rod. You directs a ray of light from a ray box into one end of the rod. The light undergoes total internal reflection at the walls of the rod and emerges at the other end. The arrangement is shown in Fig. 5.1.
The critical angle of the glass rod is 42°.
(a) Record the condition that must be met for total internal reflection to occur at the wall of the rod. [1]
(b) Calculate the refractive index of the glass. [2]
(c) Explain why the light stays inside the rod even though the rod is curved. [2]
(d) Record one advantage of sending information as light through optical fibres rather than as electrical signals through copper wires. [1]
(e) State one use of optical fibres other than telecommunications. [1]
(f) A student suggests using a glass with a critical angle of 50° for the rod. Explain whether this would work as well. [2]
(g) Describe what would happen to the light if you bends the rod too sharply. [1]
(h) State one precaution when setting up this demonstration. [1]
Answer Details
(a) For total internal reflection to occur at the wall of the rod, the angle of incidence at the wall must be greater than the critical angle (42°). [1]
Total internal reflection only happens when light travels from a denser medium (glass) into a less dense medium (air) and strikes the boundary at an angle exceeding the critical angle. Below this angle, the light refracts out of the glass instead of reflecting back in.
(b) The refractive index is related to the critical angle by: [2]
\[ n = \frac{1}{\sin c} \]
\[ n = \frac{1}{\sin 42°} = \frac{1}{0.669} = 1.49 \]
(c) Although the rod is curved, the light stays inside because at each point where it strikes the inner wall, the angle of incidence remains greater than the critical angle of 42°. [2]
The curve is gentle enough that the geometry of each successive reflection keeps the angle above 42°. Total internal reflection occurs at every reflection point along the curve, so the light bounces along inside the rod from one end to the other, following the shape of the bend.
(d) Optical fibres carry more data (higher bandwidth) than copper wires of the same diameter. [1] Other valid advantages include: less signal loss over long distances, immunity to electrical interference, and lighter, thinner cables.
(e) Endoscopes in medical imaging allow doctors to see inside the body without surgery. [1] Other valid uses include decorative lighting and illuminating hard-to-reach areas during engineering inspections.
(f) A glass with a critical angle of 50° would be less reliable for this purpose. [2]
A larger critical angle means that total internal reflection requires the light to hit the wall at a steeper angle (greater than 50° instead of 42°). With a gentle curve this might still work, but with a tighter bend, the angle of incidence at the wall could fall below 50°. At that point, light would escape through the wall by refraction instead of reflecting back in. The glass with a 42° critical angle tolerates sharper bends more reliably.
(g) If the rod is bent too sharply, the angle of incidence at the wall drops below the critical angle. [1] The light then refracts out of the rod instead of being totally internally reflected, and the signal is lost at the bend.
(h) Align the ray box carefully with the end of the rod so the light enters cleanly. [1] Other valid precautions include darkening the room to see the light path clearly, or supporting the rod so it does not roll or shift during the demonstration.
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