Lens Power Calculation for Vision Correction

This problem requires calculating the power of a lens needed to correct vision so that a person can read a book placed at 25 cm, given their minimum distance of distinct vision is 50 cm.

Understanding the Vision Problem

• The person's near point (minimum distance of distinct vision) is 50 cm. This means they can see objects clearly only if they are at or beyond 50 cm.
• The goal is to read a book at 25 cm.
• A corrective lens is needed to make the book at 25 cm appear as if it is at 50 cm, allowing the person to see it clearly. This means the lens must form a virtual image of the object (book) at the person's near point.

Applying the Lens Formula

We use the lens formula to find the focal length ($f$). The formula is:

• Object Distance ($u$): The distance of the book from the lens. Since the book is placed in front of the lens, $u = -25 \text{ cm}$.
• Image Distance ($v$): The distance where the virtual image should be formed for clear vision. This is the person's near point. Since it's a virtual image on the same side as the object, $v = -50 \text{ cm}$.

Calculating Focal Length

Substitute the values into the lens formula:

$ \frac{1}{f} = \frac{1}{(-50 \text{ cm})} - \frac{1}{(-25 \text{ cm})} $

$ \frac{1}{f} = -\frac{1}{50 \text{ cm}} + \frac{1}{25 \text{ cm}} $

Find a common denominator (50):

$ \frac{1}{f} = -\frac{1}{50 \text{ cm}} + \frac{2}{50 \text{ cm}} $

$ \frac{1}{f} = \frac{-1 + 2}{50 \text{ cm}} = \frac{1}{50 \text{ cm}} $

Thus, the focal length $f = 50 \text{ cm}$.

Calculating Lens Power

The power ($P$) of a lens is defined as the reciprocal of its focal length ($f$) in meters.

$ P = \frac{1}{f} $

First, convert the focal length to meters:

$ f = 50 \text{ cm} = 0.50 \text{ m} $

Now, calculate the power:

$ P = \frac{1}{0.50 \text{ m}} = 2 \text{ D} $

The required power of the lens is 2 Diopters.