JEE Mains Physics · Communication Systems

Line-of-Sight Range of a Tower

A space wave travels in a straight line, so the earth's curve limits its range: a tower of height h reaches d = √(2Rh), and two antennas add their ranges.

All Communication Systems notes JEE Mains Physics strategy

Why this matters

Fourteen PYQs, four of them asking for a number: five from 2021, four from 2022 and five from 2023. Seven use one antenna, finding a range, a height, an area covered or the change in range when the height changes. Seven add the ranges of a transmitting and a receiving antenna. All fourteen come from one formula, so the marks are lost on units and on adding the wrong things.

Concept 1 of 2: Range of one antenna on a round earth

From the top of a tower you can see only as far as the horizon, where your line of sight just touches the earth. The radius to that point is at right angles to the line of sight, so Pythagoras gives the distance. The tower is tiny next to the earth, so one small term can be dropped.

Definition

Line of sight touching the earth: $d^{2} = (R + h)^{2} - R^{2} = 2Rh + h^{2} \approx 2Rh$ , since $h \ll R$ .
So $d = \sqrt{2Rh}$ and $h = \dfrac{d^{2}}{2R}$ . The same formula gives the height a receiver needs when the transmitter is at ground level.
Area covered: $A = \pi d^{2} = 2\pi R h$ . Population covered = density × area.
Scaling: $d \propto \sqrt{h}$ . To double the range, the height must become 4 times as large; to triple it, 9 times.
Work in metres inside the root: $R = 6.4 \times 10^{6}$ m, and $\sqrt{2R} \approx 3578\ \text{m}^{1/2}$ .

Line-of-sight range of one antenna

d = \sqrt{2Rh}, \qquad A = \pi d^{2} = 2\pi R h

Worked example

A TV tower is 20 m high. Take

R = 6400

km. Find its range, the area it covers, and the number of people it reaches if 500 people live on each square kilometre.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 6 April 2023 · Q16Moderate

Example 1 · Communication Systems · Line-of-Sight Range of a Tower

By what percentage will the transmission range of a TV tower be affected when the height of the tower is increased by

21\%

Increased by, or increased to

To double a range the height becomes 4h, so it is increased BY 3h. Questions ask both ways, and both values are among the options.

Keep one unit inside the root

Put R and h in metres, or both in kilometres, before multiplying. A tower height in metres with R in kilometres gives a range off by a factor of about 30.

The range grows as the square root

A height 4 times as large gives a range only twice as large, because

\sqrt{4} = 2

. Scaling the range by the same factor or percentage as the height is the usual wrong option.

Drill 6 more on range of one antenna on a round earth

Concept 2 of 2: Range between a transmitting and a receiving antenna

Each antenna sees to its own horizon. The signal can get through as long as the two horizons meet, so the largest distance between the antennas is the sum of their two ranges. Add the ranges, never the heights.

Definition

Maximum line-of-sight distance: $d_M = \sqrt{2Rh_T} + \sqrt{2Rh_R} = \sqrt{2R}\left(\sqrt{h_T} + \sqrt{h_R}\right)$ .
Two identical antennas: $d_M = 2\sqrt{2Rh}$ , so $h = \dfrac{d_M^{2}}{8R}$ .
If $h_T + h_R$ is fixed, $\sqrt{h_T} + \sqrt{h_R}$ is largest when the two heights are equal. So the range is greatest for equal heights.
Heights that are 5 × a perfect square give whole-number ranges with $R = 6400$ km: 5 m → 8 km, 20 m → 16 km, 45 m → 24 km.

Line-of-sight range of two antennas

d_M = \sqrt{2Rh_T} + \sqrt{2Rh_R}

Worked example

A transmitting antenna is 405 m high and a receiving antenna is 5 m high. Take

R = 6400

km. How far apart can they be? What answer does the mistake of adding the heights first give?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 15 Apr 2023 · Q4Moderate

Example 2 · Communication Systems · Line-of-Sight Range of a Tower

The height of transmitting antenna is

180\text{ }m

and the height of the receiving antenna is

245\text{ }m

. The maximum distance between them for satisfactory communication in line of sight will be: (given

R = 6400\text{ }km

)

Add the ranges, not the heights

\sqrt{h_T} + \sqrt{h_R}

is not

\sqrt{h_T + h_R}

. Taking one root of the summed heights gives a range that is too short.

Do not forget the second antenna

When both antennas have a height, both add range. Using only the transmitter's

\sqrt{2Rh_T}

gives an answer that is always one of the options.

Identical towers divide by 8R

For two equal heights,

d = 2\sqrt{2Rh}

, so

h = d^{2}/8R

. Using

d^{2}/2R

treats the whole distance as one antenna's range and gives a height 4 times too large.

Drill 6 more on range between a transmitting and a receiving antenna

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Range of one antenna on a round earth
Line-of-sight range of one antenna
$d = \sqrt{2Rh}, \qquad A = \pi d^{2} = 2\pi R h$
Range between a transmitting and a receiving antenna
Line-of-sight range of two antennas
$d_M = \sqrt{2Rh_T} + \sqrt{2Rh_R}$

Watch out for (6)

Increased by, or increased to
→ Range of one antenna on a round earth
Keep one unit inside the root
→ Range of one antenna on a round earth
The range grows as the square root
→ Range of one antenna on a round earth
Add the ranges, not the heights
→ Range between a transmitting and a receiving antenna
Do not forget the second antenna
→ Range between a transmitting and a receiving antenna
Identical towers divide by 8R
→ Range between a transmitting and a receiving antenna

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