When a laser beam returns after reflection from an aeroplane, the observed change in frequency is `1%`, then the speed of the aeroplane is (c is the velocity of light)

Options

A. `(c)/(50)`
B. `(c)/(100)`
C. `(c)/(200)`
D. `(c)/(2)`

Answer

To solve the problem of finding the speed of the aeroplane when a laser beam returns after reflection with a 1% change in frequency, we can use the Doppler effect for light. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Doppler Effect for Light
When the laser beam is emitted towards the aeroplane, the aeroplane is moving away from the source of the light. According to the Doppler effect, the frequency observed by the moving observer (the aeroplane) will be different from the frequency emitted by the source (the laser).

### Step 2: Define the Variables
Let:
- \( f \) = original frequency of the laser beam
- \( v_0 \) = speed of the aeroplane
- \( c \) = speed of light
- The change in frequency observed after reflection is given as \( 1\% \) of \( f \).

### Step 3: Calculate the Frequency Received by the Aeroplane
When the aeroplane is moving away from the source, the frequency received by it can be calculated using the formula:
\[ f' = f \left( \frac{c - v_0}{c} \right) \]
where \( f' \) is the frequency received by the aeroplane.

### Step 4: Calculate the Frequency Reflected by the Aeroplane
After receiving the light, the aeroplane reflects it back. Now, the aeroplane acts as a source moving away from the original source. The frequency of the light reflected back is given by:
\[ f'' = f' \left( \frac{c}{c + v_0} \right) \]
Substituting \( f' \) into this equation:
\[ f'' = f \left( \frac{c - v_0}{c} \right) \left( \frac{c}{c + v_0} \right) \]

### Step 5: Set Up the Equation for Change in Frequency
The problem states that the observed change in frequency is \( 1\% \) of the original frequency:
\[ f - f'' = 0.01 f \]
This simplifies to:
\[ f - f \left( \frac{(c - v_0)c}{c(c + v_0)} \right) = 0.01 f \]
Cancelling \( f \) from both sides (assuming \( f \neq 0 \)):
\[ 1 - \frac{(c - v_0)c}{c(c + v_0)} = 0.01 \]

### Step 6: Solve for \( v_0 \)
Rearranging gives:
\[ 1 - \frac{(c - v_0)}{(c + v_0)} = 0.01 \]
This leads to:
\[ \frac{(c + v_0) - (c - v_0)}{(c + v_0)} = 0.01 \]
\[ \frac{2v_0}{(c + v_0)} = 0.01 \]
Cross-multiplying gives:
\[ 2v_0 = 0.01(c + v_0) \]
Expanding and rearranging:
\[ 2v_0 = 0.01c + 0.01v_0 \]
\[ 2v_0 - 0.01v_0 = 0.01c \]
\[ 1.99v_0 = 0.01c \]
Thus,
\[ v_0 = \frac{0.01c}{1.99} \approx \frac{c}{199} \]

### Conclusion
The speed of the aeroplane is approximately:
\[ v_0 \approx \frac{c}{200} \]