The diameter of an eye lens is `2.5xx 10^(-3)m` and the refractive index of the eye lid is `1.44`. The resolving power of the eye for light of wavelength `5000 A^(0)` will be (in `"minute"^(-1)`)

Options

A. `1.07`
B. `0.86`
C. `1.71`
D. `1.14`

Answer

To find the resolving power of the eye, we can follow these steps:

### Step 1: Understand the given values
We have the following values:
- Diameter of the eye lens, \( D = 2.5 \times 10^{-3} \, \text{m} \)
- Wavelength of light, \( \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} \)
- Refractive index of the eye lens, \( n = 1.44 \)

### Step 2: Use the formula for angular resolution
The formula for the angular resolution \( \theta \) in radians is given by:
\[ \theta = \frac{1.22 \lambda}{n D} \]

### Step 3: Substitute the values into the formula
Substituting the known values into the formula:
\[ \theta = \frac{1.22 \times (5000 \times 10^{-10})}{1.44 \times (2.5 \times 10^{-3})} \]

### Step 4: Calculate \( \theta \)
Calculating the numerator:
\[ 1.22 \times 5000 \times 10^{-10} = 6.1 \times 10^{-7} \, \text{m} \]
Calculating the denominator:
\[ 1.44 \times 2.5 \times 10^{-3} = 3.6 \times 10^{-3} \, \text{m} \]
Now substituting back:
\[ \theta = \frac{6.1 \times 10^{-7}}{3.6 \times 10^{-3}} \approx 1.69 \times 10^{-4} \, \text{radians} \]

### Step 5: Convert radians to minutes
To convert radians to minutes, we use the conversion factor \( \frac{180}{\pi} \) radians per degree and \( 60 \) minutes per degree:
\[ \text{minutes} = \theta \times \frac{180}{\pi} \times 60 \]
Substituting the value of \( \theta \):
\[ \text{minutes} = 1.69 \times 10^{-4} \times \frac{180}{\pi} \times 60 \]
Calculating this gives:
\[ \text{minutes} \approx 0.5812 \, \text{minutes} \]

### Step 6: Calculate the resolving power
The resolving power \( R \) is given by:
\[ R = \frac{1}{\theta} \]
Substituting the value of \( \theta \):
\[ R = \frac{1}{0.5812} \approx 1.71 \, \text{minute}^{-1} \]

### Final Answer
Thus, the resolving power of the eye for light of wavelength \( 5000 \, \text{Å} \) is approximately \( 1.71 \, \text{minute}^{-1} \).

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