A normal eye has retina 2 cm behind the eye-lens. What is the power of the eye-lens when the eye is (a) fully relaxed, (b) most strained?

Answer

To solve the problem, we need to find the power of the eye lens in two different conditions: when the eye is fully relaxed and when it is most strained.

### Step 1: Understanding the Problem
We know that the retina is located 2 cm behind the eye lens. We need to calculate the power of the eye lens in two scenarios:
- (a) Fully relaxed condition
- (b) Most strained condition

### Step 2: Formula for Power of a Lens
The power \( P \) of a lens is given by the formula:
\[ P = \frac{1}{f} \]
where \( f \) is the focal length in meters.

### Step 3: Fully Relaxed Condition
In the fully relaxed condition, the eye is focused on an object at infinity. Thus:
- The object distance \( u = -\infty \) (in lens formula, object distance is taken as negative)
- The image distance \( v = 2 \, \text{cm} = 0.02 \, \text{m} \)

Using the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

Substituting the values:
\[ \frac{1}{f} = \frac{1}{0.02} - \frac{1}{-\infty} \]
\[ \frac{1}{f} = 50 - 0 = 50 \, \text{D} \]

Thus, the power of the eye lens when fully relaxed is:
\[ P = 50 \, \text{D} \]

### Step 4: Most Strained Condition
In the most strained condition, the eye is focused on a nearby object. Let's assume the object distance is \( u = -25 \, \text{cm} = -0.25 \, \text{m} \). The image distance remains the same:
- Image distance \( v = 2 \, \text{cm} = 0.02 \, \text{m} \)

Using the lens formula again:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

Substituting the values:
\[ \frac{1}{f} = \frac{1}{0.02} - \frac{1}{-0.25} \]
\[ \frac{1}{f} = 50 + 4 = 54 \, \text{D} \]

Thus, the power of the eye lens when most strained is:
\[ P = 54 \, \text{D} \]

### Final Answers
- (a) The power of the eye lens when fully relaxed is **50 D**.
- (b) The power of the eye lens when most strained is **54 D**.