Question

The near and far points of a woman are 30 cm and 180 cm find power of lens she should use while reading 25 cm and by using this lens what is the maximum distance clearly visible

Answer 1

Power of lens = +0.67 D, Maximum visible distance = 81.82 cm

Answer 2

The problem involves two parts: first, calculating the power of the lens required for reading, and second, determining the maximum distance visible when using this specific lens.

Part 1: Power of the lens for reading

The woman's near point is 30 cm. This means she cannot see objects clearly if they are closer than 30 cm. She wants to read a book placed at 25 cm. To do this, the corrective lens must form a virtual image of the book (placed at 25 cm) at her natural near point (30 cm).

Given:

• Object distance, $u = -25$ cm (the book is placed 25 cm in front of the lens).
• Image distance, $v = -30$ cm (the virtual image must be formed at her near point, 30 cm in front of the lens).

Using the lens formula:

$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$

Substitute the given values:

$$\frac{1}{f} = \frac{1}{-30 \text{ cm}} - \frac{1}{-25 \text{ cm}}$$ $$\frac{1}{f} = -\frac{1}{30} + \frac{1}{25}$$

To combine the fractions, find a common denominator, which is 150:

$$\frac{1}{f} = \frac{-5}{150} + \frac{6}{150}$$ $$\frac{1}{f} = \frac{1}{150}$$

So, the focal length of the lens is $f = +150$ cm. To find the power of the lens, convert the focal length to meters: $f = 150 \text{ cm} = 1.5 \text{ m}$ The power of the lens ($P$) is given by:

$$P = \frac{1}{f \text{ (in meters)}}$$ $$P = \frac{1}{1.5} = \frac{10}{15} = \frac{2}{3} \text{ Diopters (D)}$$ $$P \approx +0.67 \text{ D}$$

Since the power is positive, it is a converging (convex) lens, which is expected for correcting presbyopia/hypermetropia.

Part 2: Maximum distance clearly visible using this lens

The woman's natural far point is 180 cm. This means without any corrective lens, she can see objects clearly up to 180 cm. Now, she is using the reading lens found in Part 1 (with focal length $f = +150$ cm). We need to find the maximum distance an object can be placed such that she can see it clearly through this lens.

For her to see an object clearly, the lens must form an image of that object at or within her natural far point. For the maximum distance, the image should be formed at her far point. Given:

• Focal length of the lens, $f = +150$ cm.
• Image distance, $v = -180$ cm (the virtual image must be formed at her far point, 180 cm in front of the lens).
• We need to find the object distance, $u$.

Using the lens formula:

$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$

Substitute the values:

$$\frac{1}{150} = \frac{1}{-180} - \frac{1}{u}$$

Rearrange to solve for $u$:

$$\frac{1}{u} = \frac{1}{-180} - \frac{1}{150}$$ $$\frac{1}{u} = -\left(\frac{1}{180} + \frac{1}{150}\right)$$

To combine the fractions, find a common denominator, which is 900:

$$\frac{1}{u} = -\left(\frac{5}{900} + \frac{6}{900}\right)$$ $$\frac{1}{u} = -\frac{11}{900}$$

So, the object distance is:

$$u = -\frac{900}{11} \text{ cm}$$

The maximum distance clearly visible is the absolute value of $u$:

$$\text{Maximum distance} = \frac{900}{11} \text{ cm} \approx 81.82 \text{ cm}$$

This means that when wearing her reading glasses, she can only see objects clearly up to approximately 81.82 cm. This is consistent with the fact that convex lenses used for reading can impair distant vision.