The head lights of a jeep are 1.2 m apart. If the pupil of the eye of an observer has a diameter of 2 mm and light of wavelength 5896 Å is used, what should be the maximum distance of the jeep from the observer if the two head lights are just separated

Options

A. 33.9 km
B. 33.9 m
C. 3.39 km
D. 3.39 m

Explanation

To solve the problem, we need to determine the maximum distance \( x \) from the observer to the jeep, where the two headlights are just separated. We will use the principles of diffraction and the formula for angular resolution.

### Step-by-Step Solution:

1. **Identify the given values:**
- Distance between the headlights, \( D = 1.2 \, \text{m} \)
- Diameter of the pupil of the eye, \( d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
- Wavelength of light, \( \lambda = 5896 \, \text{Å} = 5896 \times 10^{-10} \, \text{m} \)

2. **Understanding the angular resolution:**
The condition for the two headlights to be just resolved (separated) is given by the Rayleigh criterion:
\[ \theta = \frac{1.22 \lambda}{d} \]
where \( \theta \) is the angular separation.

3. **Relate the angle to the distance:**
The angle \( \theta \) can also be expressed in terms of the distance \( x \) from the observer:
\[ \theta = \frac{D}{x} \]

4. **Set the equations equal:**
Since both expressions represent the same angle \( \theta \), we can set them equal to each other:
\[ \frac{1.22 \lambda}{d} = \frac{D}{x} \]

5. **Rearranging to find \( x \):**
Rearranging the equation gives:
\[ x = \frac{D \cdot d}{1.22 \lambda} \]

6. **Substituting the values:**
Now substitute the values into the equation:
\[ x = \frac{1.2 \, \text{m} \cdot (2 \times 10^{-3} \, \text{m})}{1.22 \cdot (5896 \times 10^{-10} \, \text{m})} \]

7. **Calculating the numerator and denominator:**
- Numerator: \( 1.2 \times 2 \times 10^{-3} = 2.4 \times 10^{-3} \, \text{m}^2 \)
- Denominator: \( 1.22 \times 5896 \times 10^{-10} \approx 7.194 \times 10^{-7} \, \text{m} \)

8. **Calculating \( x \):**
\[ x = \frac{2.4 \times 10^{-3}}{7.194 \times 10^{-7}} \approx 3336.5 \, \text{m} \]

9. **Converting to kilometers:**
\[ x \approx 3.3365 \, \text{km} \approx 3.34 \, \text{km} \]

10. **Final answer:**
Rounding to two decimal places, the maximum distance of the jeep from the observer is approximately:
\[ x \approx 3.34 \, \text{km} \]