The near point and far point of child are at 10cm and 100cm.If the retina is 2cm behind the eyelens,what is the range of power of eyelens

To determine the range of power of the eye lens for a child with a near point of 10 cm and a far point of 100 cm, we can use the lens formula and the concept of power in optics. Let's break this down step by step.

Understanding the Eye's Optical System

The eye functions similarly to a camera, where the lens focuses light onto the retina. The distance from the lens to the retina is crucial for calculating the lens's power. In this case, the distance from the lens to the retina is given as 2 cm.

Key Definitions

• Near Point: The closest distance at which the eye can focus on an object, which is 10 cm in this case.
• Far Point: The farthest distance at which the eye can see an object clearly, which is 100 cm here.
• Power of a Lens (P): Defined as P = 1/f, where f is the focal length in meters.

Calculating Focal Lengths

To find the power of the eye lens, we first need to calculate the focal lengths for both the near point and the far point using the lens formula:

1. **For the Near Point (10 cm):**

Using the lens formula, we can express it as:

\( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)

Where:

• v = distance from the lens to the retina = -2 cm (negative because it is on the same side as the object)
• u = distance to the near point = -10 cm (also negative for the same reason)

Substituting the values:

\( \frac{1}{f} = \frac{1}{-2} - \frac{1}{-10} \)

\( \frac{1}{f} = -\frac{1}{2} + \frac{1}{10} = -\frac{5}{10} + \frac{1}{10} = -\frac{4}{10} = -0.4 \)

Thus, \( f = -2.5 \) cm or -0.025 m.

2. **For the Far Point (100 cm):**

Using the same lens formula:

Here, u = -100 cm:

\( \frac{1}{f} = \frac{1}{-2} - \frac{1}{-100} \)

\( \frac{1}{f} = -\frac{1}{2} + \frac{1}{100} = -\frac{50}{100} + \frac{1}{100} = -\frac{49}{100} \)

Thus, \( f = -2.04 \) cm or -0.0204 m.

Calculating the Power of the Lens

Now that we have the focal lengths, we can calculate the power of the lens:

1. **Power for Near Point:**

Using the formula \( P = \frac{1}{f} \):

For \( f = -0.025 \) m:

Power \( P_{near} = \frac{1}{-0.025} = -40 \) diopters.

2. **Power for Far Point:**

For \( f = -0.0204 \) m:

Power \( P_{far} = \frac{1}{-0.0204} \approx -49 \) diopters.

Range of Power of the Eye Lens

The range of power of the eye lens for this child is from approximately -40 diopters (for the near point) to -49 diopters (for the far point). This means the eye lens must be capable of adjusting its power within this range to focus on objects at varying distances effectively.

In summary, the eye lens of the child has a power range of about -40 to -49 diopters, allowing for clear vision from 10 cm to 100 cm. This range is essential for accommodating different distances, ensuring that the child can see both near and far objects clearly.