JEE MainPhysics

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Propagation Velocity and Echo Time of Electromagnetic Waves

Q: Why is the speed of light not 3 × 10^8m/s?

The problem states the wave travels in a medium. In any medium, the speed of an electromagnetic wave is related to the field amplitudes by v = E_0 / B_0, which evaluates here to 1.5 × 10^8m/s.

Q: Why do we multiply the distance by 2?

A radar echo travels to the target and then reflects back to the source, covering twice the line-of-sight distance (2d).

A radar sends an electromagnetic signal of electric field (E_0) = 2.25V/m and magnetic field (B_0) = 1.5 × 10^(-8)T which strikes a target on line of sight at a distance of 3km in a medium. After that, a pail of signal (echo) reflects back towards the radar vit1i same velocity and by same path. If the signal was transmitted at time t_0 from radar. then after how much time echo will reach to the radar?

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A2.0 × 10^(-5)sB4.0 × 10^(-5)sC1.0 × 10^(-5)sD8.0 × 10^(-5)s

Step-by-step solutionView

Correct answer

The electromagnetic wave travels at speed $v = \frac{E_{0}}{B_{0}} = 1.5 \times 10^{8} m / s$ in the medium, requiring $4.0 \times 10^{- 5} s$ to travel the total round-trip distance of $6 k m$ .

Option analysis

Why each option works or fails

A · 2.0 × 10^(-5)s

The student calculates only the one-way travel time of the signal from the radar to the target. Recall that an echo requires the signal to travel to the target and back, meaning the total distance traveled is $2 d$ instead of $d$ .

B · 4.0 × 10^(-5)s

The wave speed is calculated via $v = E_{0} / B_{0} = 1.5 \times 10^{8} m / s$ . The round-trip distance is $2 d = 6 \times 10^{3} m$ . The total time is $t = 2 d / v = 4.0 \times 10^{- 5} s$ .

C · 1.0 × 10^(-5)s

The student incorrectly assumes the wave travels at the speed of light in a vacuum ( $c = 3 \times 10^{8} m / s$ ) instead of calculating the actual velocity in the medium, and only considers the one-way trip. The velocity of an electromagnetic wave in a medium is determined by the ratio of the amplitudes of its electric and magnetic fields, $v = \frac{E_{0}}{B_{0}}$ , which may differ from $c$ .

D · 8.0 × 10^(-5)s

The student makes an arithmetic error during the division or multiplication of powers of ten. Carefully track the exponent signs when dividing by a velocity containing a positive power of ten: $\frac{6 \times 10^{3}}{1.5 \times 10^{8}} = 4 \times 10^{- 5} s$ .

Reviewed route

Solution

StepWorking

01Given

$E_{0} = 2.25 V/m$ , $B_{0} = 1.5 \times 10^{- 8} T$ , target distance $d = 3 km = 3 \times 10^{3} m$ .

02Find

Total round-trip time $Δ t$ for the radar echo to return.

03Strategise

The speed of the electromagnetic wave in the medium is given by $v = \frac{E_{0}}{B_{0}}$ . The total distance travelled for the echo is $2 d$ . Thus, the total time is $Δ t = \frac{2 d}{v}$ .

04Execute

Calculate the speed in the medium: $v = \frac{2.25}{1.5 \times 10^{- 8}} = 1.5 \times 10^{8} m/s$ Then calculate the round-trip time: $Δ t = \frac{2 \times 3000}{1.5 \times 10^{8}} = 4.0 \times 10^{- 5} s$

✓Verify

Units are seconds. Speed in medium $1.5 \times 10^{8} m/s < c$ , which is physically valid for a dielectric medium with refractive index $n = 2$ . Round-trip time is twice the one-way time ( $2.0 \times 10^{- 5} s$ ).

Hints that build this answer step by step

How do we find the velocity $v$ of the electromagnetic wave in this medium using the given amplitudes of the electric field ( $E_{0}$ ) and magnetic field ( $B_{0}$ )?
$v = \frac{E_{0}}{B_{0}}$
Calculate the velocity $v$ of the electromagnetic wave in the medium.
$1.5 \times 10^{8} m / s$
What is the total distance the wave must travel to go to the target and return to the radar?
$6 k m$
Calculate the total time $t$ elapsed for the echo to return.
$4.0 \times 10^{- 5} s$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Physics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 22 September 2026

Quick checks

Students also ask

Why is the speed of light not

3 \times 10^{8} m/s

The problem states the wave travels in a medium. In any medium, the speed of an electromagnetic wave is related to the field amplitudes by $v = E_{0} / B_{0}$ , which evaluates here to $1.5 \times 10^{8} m/s$ .

Why do we multiply the distance by 2?

A radar echo travels to the target and then reflects back to the source, covering twice the line-of-sight distance ( $2 d$ ).

Answer

The electromagnetic wave travels at speed $v = \frac{E_{0}}{B_{0}} = 1.5 \times 10^{8} m / s$ in the medium, requiring $4.0 \times 10^{- 5} s$ to travel the total round-trip distance of $6 k m$ .

Why each option works or fails

A: 2.0 × 10^(-5)s - The student calculates only the one-way travel time of the signal from the radar to the target. Recall that an echo requires the signal to travel to the target and back, meaning the total distance traveled is $2 d$ instead of $d$ .
B · correct: 4.0 × 10^(-5)s - The wave speed is calculated via $v = E_{0} / B_{0} = 1.5 \times 10^{8} m / s$ . The round-trip distance is $2 d = 6 \times 10^{3} m$ . The total time is $t = 2 d / v = 4.0 \times 10^{- 5} s$ .
C: 1.0 × 10^(-5)s - The student incorrectly assumes the wave travels at the speed of light in a vacuum ( $c = 3 \times 10^{8} m / s$ ) instead of calculating the actual velocity in the medium, and only considers the one-way trip. The velocity of an electromagnetic wave in a medium is determined by the ratio of the amplitudes of its electric and magnetic fields, $v = \frac{E_{0}}{B_{0}}$ , which may differ from $c$ .
D: 8.0 × 10^(-5)s - The student makes an arithmetic error during the division or multiplication of powers of ten. Carefully track the exponent signs when dividing by a velocity containing a positive power of ten: $\frac{6 \times 10^{3}}{1.5 \times 10^{8}} = 4 \times 10^{- 5} s$ .

Step-by-step solution

Given: $E_{0} = 2.25 V/m$ , $B_{0} = 1.5 \times 10^{- 8} T$ , target distance $d = 3 km = 3 \times 10^{3} m$ .
Find: Total round-trip time $Δ t$ for the radar echo to return.
Strategise: The speed of the electromagnetic wave in the medium is given by $v = \frac{E_{0}}{B_{0}}$ . The total distance travelled for the echo is $2 d$ . Thus, the total time is $Δ t = \frac{2 d}{v}$ .
Execute: Calculate the speed in the medium: $v = \frac{2.25}{1.5 \times 10^{- 8}} = 1.5 \times 10^{8} m/s$ Then calculate the round-trip time: $Δ t = \frac{2 \times 3000}{1.5 \times 10^{8}} = 4.0 \times 10^{- 5} s$
Verify: Units are seconds. Speed in medium $1.5 \times 10^{8} m/s < c$ , which is physically valid for a dielectric medium with refractive index $n = 2$ . Round-trip time is twice the one-way time ( $2.0 \times 10^{- 5} s$ ).

Propagation Velocity and Echo Time of Electromagnetic Waves

What feels right?

The electromagnetic wave travels at speed $v = \frac{E_{0}}{B_{0}} = 1.5 \times 10^{8} m / s$ in the medium, requiring $4.0 \times 10^{- 5} s$ to travel the total round-trip distance of $6 k m$ .

Why each option works or fails

Solution

01Given

02Find

03Strategise

04Execute

✓Verify

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The electromagnetic wave travels at speed $v = \frac{E_{0}}{B_{0}} = 1.5 \times 10^{8} m / s$ in the medium, requiring $4.0 \times 10^{- 5} s$ to travel the total round-trip distance of $6 k m$ .

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