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 dots can be resolved by the eye? [Take wavelelngth of light =500nm]

Options

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

Answer

To solve the problem of determining the maximum distance at which two point white dots can be resolved by the human eye, we can use the formula for the resolving power of an optical system. The key steps are as follows:

### Step 1: Understand the Parameters
We have:
- Distance between the dots, \( d = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} \)
- Diameter of the pupil, \( Y = 3 \text{ mm} = 3 \times 10^{-3} \text{ m} \)
- Wavelength of light, \( \lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m} \)

### Step 2: Use the Resolving Power Formula
The resolving power of the eye can be expressed as:
\[ \frac{y}{d} \geq \frac{1.22 \lambda}{D} \]
Where:
- \( y \) is the diameter of the pupil,
- \( d \) is the distance between the two dots,
- \( D \) is the maximum distance at which the dots can be resolved.

Rearranging the formula to find \( D \):
\[ D \leq \frac{d \cdot Y}{1.22 \lambda} \]

### Step 3: Substitute the Values
Substituting the known values into the equation:
\[ D \leq \frac{(1 \times 10^{-3} \text{ m}) \cdot (3 \times 10^{-3} \text{ m})}{1.22 \cdot (500 \times 10^{-9} \text{ m})} \]

### Step 4: Calculate the Denominator
Calculating the denominator:
\[ 1.22 \cdot (500 \times 10^{-9}) = 6.1 \times 10^{-7} \text{ m} \]

### Step 5: Calculate the Numerator
Calculating the numerator:
\[ (1 \times 10^{-3}) \cdot (3 \times 10^{-3}) = 3 \times 10^{-6} \text{ m}^2 \]

### Step 6: Final Calculation
Now, substituting back into the equation for \( D \):
\[ D \leq \frac{3 \times 10^{-6}}{6.1 \times 10^{-7}} \approx 4.92 \text{ m} \]

### Step 7: Conclusion
Thus, the maximum distance \( D \) at which the two dots can be resolved by the eye is approximately:
\[ D \approx 5 \text{ m} \]