For a normal eye, the cornea of eye provides a converging power of `40 D` and the least converging power of the eye lens behind the cornea is `20 D`. Using this information, the distance between the retina and the cornea eye lens can be estimated to be

Options

A. 5cm
B. 2.5cm
C. 1.67cm
D. 1.5cm

Explanation

To solve the problem, we need to find the distance between the retina and the cornea-eye lens using the given converging powers of the cornea and the eye lens.

### Step-by-Step Solution:

1. **Identify the Powers of the Lenses**:
- The power of the cornea (P1) = 40 D (diopters)
- The power of the eye lens (P2) = 20 D (diopters)

2. **Calculate the Equivalent Power**:
- Since both the cornea and the eye lens are converging lenses, their powers add up.
- Equivalent Power (P_eq) = P1 + P2
- P_eq = 40 D + 20 D = 60 D

3. **Calculate the Focal Length**:
- The focal length (f) of a lens is related to its power (P) by the formula:
\[ P = \frac{1}{f} \]
- Rearranging gives:
\[ f = \frac{1}{P} \]
- For the equivalent power:
\[ f_{eq} = \frac{1}{60} \text{ meters} \]

4. **Convert Focal Length to Centimeters**:
- To convert meters to centimeters, multiply by 100:
\[ f_{eq} = \frac{1}{60} \times 100 = \frac{100}{60} \text{ cm} = \frac{10}{6} \text{ cm} = \frac{5}{3} \text{ cm} \]
- This simplifies to approximately:
\[ f_{eq} \approx 1.67 \text{ cm} \]

5. **Interpret the Result**:
- The focal length of 1.67 cm indicates that for an object at infinity, the image is formed at the retina, which is approximately 1.67 cm from the cornea-eye lens.

### Final Answer:
The distance between the retina and the cornea-eye lens is approximately **1.67 cm**.

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