Which wireless medium cannot pass through walls and is therefore used in television remote controls?

The correct answer is D, transparent plastic. An optical fibre carries information as pulses of light, not as an electric current, so its core and cladding must be transparent and not conducting. The two materials used are ultra-pure silica glass for long-distance cables and transparent plastic, usually an acrylic polymer, for short cheap links; among the four options only transparent plastic can pass light. Option A, copper, is the conductor of twisted pair and coaxial cable, which carry electrical signals and are exactly what fibre replaced. Option B, aluminium, is used for overhead power lines and for shielding, and like any metal it is opaque. Option C, steel, appears only as strength members in the outer sheath of some cables, never as the light-carrying core. The examinable point is that an optical fibre is a dielectric, not a metal.

The correct answer is B, sources. At the transmitting end of a fibre link the electrical signal must be turned into light, and the two devices that do it are the light-emitting diode and the laser diode. The light-emitting diode is cheap, long-lived and spreads its light over a range of wavelengths, so it suits short multi-mode links; the laser diode gives a narrow, coherent beam that couples into a thin single-mode core and so suits long high-speed routes. Option A, detectors, is the job of the photodiode at the far end, which turns the arriving light back into current. Option C, repeaters, describes a complete unit placed along a long route to receive, clean and resend the signal; a laser may sit inside one, but a repeater is not what the diode itself is called. Option D, amplifiers, strengthen a signal, and in fibre that work is done by an erbium-doped fibre amplifier, not by the source.

The correct answer is B, 540 kHz. In frequency division multiplexing each channel is given its own slice of the frequency band, and a small unused strip called a guard band is left between two neighbouring channels so that one does not spill into the next. Five channels of 100 kHz each need 500 kHz. Guard bands are needed only in the gaps between channels, and five channels placed side by side have four gaps, so four guard bands of 10 kHz add 40 kHz. The total is 500 plus 40, that is 540 kHz. Option A, 500 kHz, is the answer of a candidate who forgot the guard bands altogether. Option D, 560 kHz, comes from using five guard bands instead of four, the classic fencepost mistake of counting gaps as if they equalled the number of items. Option C, 460 kHz, wrongly subtracts the guard bands instead of adding them. Remember the rule: for n channels there are n minus 1 guard bands.

The correct answer is B, total internal reflection. The fibre has a core of higher refractive index surrounded by a cladding of lower refractive index. Light that enters the core and meets the boundary at an angle greater than the critical angle is reflected entirely back into the core, with none of it refracted away into the cladding. Repeating this thousands of times, the ray travels along the fibre even around bends. Option A, diffraction, is the bending of a wave around an obstacle or through a narrow slit and plays no part here. Option C, polarisation, is the restriction of a wave vibration to one plane, used in sunglasses and in liquid crystal displays, not in guiding light along a fibre. Option D, dispersion, is the splitting of light into its colours and is in fact an unwanted effect inside a fibre, since it spreads a pulse out and limits the data rate.

The correct answer is C, optical fibre. Electromagnetic interference acts on moving charge, so any medium that carries a signal as an electric current can pick up noise from motors, fluorescent lamps, lightning and nearby cables. An optical fibre carries the signal as light through glass or plastic and no current flows in it at all, so stray fields cannot add anything to the signal. Option A, unshielded twisted pair, is the most vulnerable of all the guided media; twisting the pair cancels much of the crosstalk but leaves it open to outside fields. Option B, coaxial cable, is better protected because of its braided metal shield, yet it still carries current and is not immune. Option D, radio waves, is the opposite case altogether, since the signal travels through open space and shares the band with every other transmitter. This immunity, with its huge bandwidth, is why fibre carries the trunk routes.