The near point and the far point of a child are at 10 cm and 100 cm. If the retina is 2.0 cm behind the eye-lens, what is the range of the power of the eye-lens?

Answer

To solve the problem, we need to determine the range of the power of the eye lens for a child with a near point of 10 cm and a far point of 100 cm, given that the retina is located 2.0 cm behind the eye lens.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Near Point (N) = 10 cm = -0.1 m (u = -10 cm, negative because it is a virtual object)
- Far Point (F) = 100 cm = -1 m (u = -100 cm, negative for the same reason)
- Distance from the lens to the retina (v) = 2 cm = 0.02 m (positive because it is a real image)

2. **Calculate the Power for the Near Point:**
- For the near point, we use the lens formula:
\[ \frac{1}{F} = \frac{1}{v} - \frac{1}{u} \]
- Substituting the values:
\[ \frac{1}{F} = \frac{1}{0.02} - \frac{1}{-0.1} \]
\[ \frac{1}{F} = 50 + 10 = 60 \text{ m}^{-1} \]
- The focal length (F) is:
\[ F = \frac{1}{60} \text{ m} \approx 0.01667 \text{ m} \]
- The power (P) of the lens is:
\[ P = \frac{1}{F} = 60 \text{ diopters} \]

3. **Calculate the Power for the Far Point:**
- For the far point, we again use the lens formula:
\[ \frac{1}{F} = \frac{1}{v} - \frac{1}{u} \]
- Substituting the values:
\[ \frac{1}{F} = \frac{1}{0.02} - \frac{1}{-1} \]
\[ \frac{1}{F} = 50 + 1 = 51 \text{ m}^{-1} \]
- The focal length (F) is:
\[ F = \frac{1}{51} \text{ m} \approx 0.01961 \text{ m} \]
- The power (P) of the lens is:
\[ P = \frac{1}{F} = 51 \text{ diopters} \]

4. **Determine the Range of Power:**
- The range of the power of the eye lens is from the power calculated for the far point to the power calculated for the near point:
\[ \text{Range of Power} = 51 \text{ D} \text{ to } 60 \text{ D} \]

### Final Answer:
The range of the power of the eye lens is from **51 diopters to 60 diopters**.