Radio waves of frequency 600 MHz are sent by a radar towards an enemy aircraft. The frequency of the radio waves reflected from the aircraft as measured at the radar station is found to increase by 6 kHz. It follows that the aircraft is :

Options

A. Approaching the radar station with a speed `1.5 kms^(-1)`
B. Going away from the radar station with a speed `1.5 kms^(-1)`
C. Approaching the radar station with a speed `3kms^(-1)`
D. Going away from the radar station with a speed `3kms^(-1)`

Explanation

To solve the problem, we need to determine whether the aircraft is approaching or moving away from the radar based on the frequency shift of the radio waves. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Information
- The frequency of the radio waves sent by the radar (N₀) = 600 MHz = 600 × 10⁶ Hz.
- The increase in frequency (ΔN) of the reflected waves as measured at the radar station = 6 kHz = 6 × 10³ Hz.

### Step 2: Use the Doppler Effect Formula
For a source of sound (or electromagnetic waves) and a moving observer, the observed frequency (Nᵣ) can be calculated using the Doppler effect formula:
\[ Nᵣ = N₀ \left( \frac{v_s + v_a}{v_s - v_a} \right) \]
where:
- \( v_s \) = speed of the waves in the medium (for radio waves, this is the speed of light, \( c \approx 3 \times 10^8 \) m/s),
- \( v_a \) = speed of the observer (aircraft in this case).

### Step 3: Calculate the Change in Frequency
The change in frequency (ΔN) can be expressed as:
\[ ΔN = Nᵣ - N₀ \]
Substituting the expression for \( Nᵣ \):
\[ ΔN = N₀ \left( \frac{v_s + v_a}{v_s - v_a} \right) - N₀ \]
\[ ΔN = N₀ \left( \frac{(v_s + v_a) - (v_s - v_a)}{v_s - v_a} \right) \]
\[ ΔN = N₀ \left( \frac{2v_a}{v_s - v_a} \right) \]

### Step 4: Rearranging the Equation
From the above equation, we can isolate \( v_a \):
\[ v_a = \frac{ΔN (v_s - v_a)}{2N₀} \]
This leads to:
\[ v_a = \frac{ΔN v_s}{2N₀ + ΔN} \]

### Step 5: Substitute the Values
Now, substituting the known values:
- \( ΔN = 6 \times 10^3 \) Hz,
- \( N₀ = 600 \times 10^6 \) Hz,
- \( v_s = 3 \times 10^8 \) m/s.

Substituting these into the equation:
\[ v_a = \frac{(6 \times 10^3) (3 \times 10^8)}{2(600 \times 10^6) + 6 \times 10^3} \]

### Step 6: Simplifying the Equation
Calculating the denominator:
\[ 2(600 \times 10^6) + 6 \times 10^3 \approx 1200 \times 10^6 \]
Thus,
\[ v_a \approx \frac{(6 \times 10^3) (3 \times 10^8)}{1200 \times 10^6} \]
\[ v_a \approx \frac{18 \times 10^{11}}{1200 \times 10^6} \]
\[ v_a \approx 1.5 \times 10^3 \text{ m/s} \]
or
\[ v_a \approx 1.5 \text{ km/s} \]

### Step 7: Conclusion
Since the frequency of the reflected waves has increased, it indicates that the aircraft is approaching the radar. Therefore, the aircraft is approaching the radar with a speed of 1.5 km/s.