Two point white dots are 1mm apart on a black paper. They are viewed by eye of pupil diameter 3mm. Approximately, what is the maximum distance at which these dits can be resolved by the eye? [Take wavelelngth of light =500nm]

Options

A. 1m
B. 5m
C. 3m
D. 6m

Explanation

To solve the problem of determining the maximum distance at which two point white dots, 1 mm apart, can be resolved by the eye with a pupil diameter of 3 mm, we can use the Rayleigh criterion for resolution. The formula we will use is:

\[ D = \frac{y \cdot d}{1.22 \lambda} \]

Where:
- \( D \) = maximum distance at which the dots can be resolved
- \( y \) = distance between the two dots (1 mm = \( 1 \times 10^{-3} \) m)
- \( d \) = diameter of the pupil (3 mm = \( 3 \times 10^{-3} \) m)
- \( \lambda \) = wavelength of light (500 nm = \( 500 \times 10^{-9} \) m)

### Step-by-Step Solution:

1. **Identify the parameters:**
- Distance between dots, \( y = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Diameter of the pupil, \( d = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m} \)
- Wavelength of light, \( \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \)

2. **Substitute the values into the formula:**
\[ D = \frac{(1 \times 10^{-3} \, \text{m}) \cdot (3 \times 10^{-3} \, \text{m})}{1.22 \cdot (500 \times 10^{-9} \, \text{m})} \]

3. **Calculate the numerator:**
\[ \text{Numerator} = 1 \times 10^{-3} \cdot 3 \times 10^{-3} = 3 \times 10^{-6} \, \text{m}^2 \]

4. **Calculate the denominator:**
\[ \text{Denominator} = 1.22 \cdot 500 \times 10^{-9} = 6.1 \times 10^{-7} \, \text{m} \]

5. **Now, calculate \( D \):**
\[ D = \frac{3 \times 10^{-6}}{6.1 \times 10^{-7}} \approx 4.92 \, \text{m} \]

6. **Round the result:**
\[ D \approx 5 \, \text{m} \]

### Final Answer:
The maximum distance at which the two dots can be resolved by the eye is approximately **5 meters**.