Human eye is sensitive to light of wavelength `5,500 Å`. Calculate the limit resolution of eye if diameter of pupil in the eye is nearly 2 mm.

Answer

To calculate the limit of resolution of the human eye, we can use the formula for angular resolution given by:

\[ \theta = \frac{1.22 \lambda}{d} \]

where:
- \(\theta\) is the limit of resolution (in radians),
- \(\lambda\) is the wavelength of light (in meters),
- \(d\) is the diameter of the pupil (in meters).

### Step 1: Convert the wavelength from angstroms to meters
The wavelength given is \(5500 \, \text{Å}\). We know that:
\[ 1 \, \text{Å} = 10^{-10} \, \text{m} \]
Thus,
\[ \lambda = 5500 \, \text{Å} = 5500 \times 10^{-10} \, \text{m} = 5.5 \times 10^{-7} \, \text{m} \]

### Step 2: Convert the diameter of the pupil from mm to meters
The diameter of the pupil is given as \(2 \, \text{mm}\). We know that:
\[ 1 \, \text{mm} = 10^{-3} \, \text{m} \]
Thus,
\[ d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \]

### Step 3: Substitute the values into the resolution formula
Now we can substitute the values of \(\lambda\) and \(d\) into the formula for \(\theta\):
\[ \theta = \frac{1.22 \times (5.5 \times 10^{-7})}{2 \times 10^{-3}} \]

### Step 4: Calculate the limit of resolution
Calculating the above expression:
\[ \theta = \frac{1.22 \times 5.5 \times 10^{-7}}{2 \times 10^{-3}} = \frac{6.71 \times 10^{-7}}{2 \times 10^{-3}} = 3.355 \times 10^{-4} \, \text{radians} \]

### Step 5: Round the result
Rounding \(3.355 \times 10^{-4}\) gives us:
\[ \theta \approx 3.36 \times 10^{-4} \, \text{radians} \]

Thus, the limit of resolution of the human eye is approximately \(3.36 \times 10^{-4} \, \text{radians}\).

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