A transmitting antenna at the top of a tower has a height 18 m and the height of the receiving antenna is 32 m. What is the maximum distance between them for satisfactory communication in line-of-sight mode? Given radius of earth is \( 6.4\times10^{6}\,\mathrm{m} \).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 35.4 km
In line-of-sight communication the curvature of the earth limits the range. An antenna of height \( h \) can reach up to a distance \( \sqrt{2Rh} \) along the earth's surface (its radio horizon). The maximum distance between two antennas is the sum of their two horizons.
\( \displaystyle d_m = \sqrt{2Rh_T} + \sqrt{2Rh_R} \)
\( 2R = 2 \times 6.4\times10^{6} = 1.28\times10^{7}\,\mathrm{m} \)
\( \displaystyle \sqrt{2Rh_T} = \sqrt{1.28\times10^{7} \times 18} = \sqrt{2.304\times10^{8}} = 1.518\times10^{4}\,\mathrm{m} \)
\( \displaystyle \sqrt{2Rh_R} = \sqrt{1.28\times10^{7} \times 32} = \sqrt{4.096\times10^{8}} = 2.024\times10^{4}\,\mathrm{m} \)
\( \displaystyle d_m = 1.518\times10^{4} + 2.024\times10^{4} = 3.54\times10^{4}\,\mathrm{m} = 35.4\,\mathrm{km} \)
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Space Wave – “… If the transmitting antenna is at a height hT, then you can show that the distance to the horizon dT is given as
\( \displaystyle d_T = \sqrt{2Rh_T} \)
where R is the radius of the earth (approximately 6400 km). … With reference to Fig. 15.5 the maximum line-of-sight distance dM between the two antennas having heights hT and hR above the earth is given by
\( \displaystyle d_M = \sqrt{2Rh_T} + \sqrt{2Rh_R} \qquad (15.1) \)
where hR is the height of receiving antenna.”