Question

If the least distance of distinct vision for a boy is \( 35 \, \text{cm} \), then the lens to be used by the boy for correcting the defect of his eye is

• A. Convex lens of focal length \( 35 \, \text{cm} \)
• B. Concave lens of focal length \( 35 \, \text{cm} \)
• C. Convex lens of focal length \( 87.5 \, \text{cm} \)
• D. Concave lens of focal length \( 87.5 \, \text{cm} \)

Hint:

For vision defects, recall the normal human eye's near point (\( 25 cm} \)) and far point (infinity). Myopia (nearsightedness) means the far point is closer than infinity, corrected by a concave lens. Hypermetropia (farsightedness) means the near point is farther than \( 25 cm} \), corrected by a convex lens. Presbyopia is similar to hypermetropia, affecting older individuals. Always use the lens formula (\( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)) with appropriate sign conventions for object distance (\( u \)) and image distance (\( v \)). For correcting vision, the object is placed at the normal near point (for hypermetropia) or at infinity (for myopia), and the image is formed at the person's defective near/far point.

Answer 1

The Correct Option is C

Step 1: Identify the eye defect.
The normal least distance of distinct vision (near point) is \( 25 \, \text{cm} \). For the given boy, the least distance of distinct vision is \( 35 \, \text{cm} \). This means he cannot see objects clearly closer than \( 35 \, \text{cm} \). This defect is hypermetropia (farsightedness), where the eye lens converges light too weakly, causing the image to form behind the retina.
Step 2: Determine the type of lens required.
To correct hypermetropia, a convex lens is used. A convex lens converges the light rays before they enter the eye, allowing the eye's lens to focus the image correctly on the retina.
Step 3: Calculate the focal length of the corrective lens.
For a hypermetropic eye, the lens needs to form a virtual image of an object placed at the normal near point (\( u = -25 \, \text{cm} \)) at the person's defective near point (\( v = -35 \, \text{cm} \)). We use the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Here, \( u = -25 \, \text{cm} \) (object at normal near point)
\( v = -35 \, \text{cm} \) (virtual image formed at the defective near point)
Substitute the values into the lens formula:
\[ \frac{1}{f} = \frac{1}{-35} - \frac{1}{-25} \]
\[ \frac{1}{f} = -\frac{1}{35} + \frac{1}{25} \]
To add these fractions, find a common denominator, which is \( 175 \).
\[ \frac{1}{f} = \frac{-5}{175} + \frac{7}{175} \]
\[ \frac{1}{f} = \frac{7 - 5}{175} \]
\[ \frac{1}{f} = \frac{2}{175} \]
Therefore, the focal length \( f \) is:
\[ f = \frac{175}{2} \, \text{cm} \]
\[ f = 87.5 \, \text{cm} \]
Since the focal length is positive, it confirms that a convex lens is required.