A person has a relaxed eye (infinite far point) with a focal length of $2.5\text{cm}$. The eye can accommodate an additional $20\text{D}$ of power. Calculate the least distance of distinct vision (near point) for this person, in centimeters.

Step-by-step solution

1. The power of the relaxed eye is $P_{relaxed}=\frac{1}{f}=\frac{1}{0.025\text{m}}=40\text{D}$.
2. Maximum accommodated power is $P_{max}=P_{relaxed}+20=60\text{D}$.
3. When the eye focuses on the near point, the image is formed at the retina, i.e., at the focal length of the eye. Hence $\frac{1}{v}=P_{max}$ with $v$ measured from the lens to the retina (which is the focal length).
4. The object distance $u$ (the near point) satisfies the lens formula $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$. Here $f=2.5\text{cm}$, $\frac{1}{v}=P_{max}=60\text{D}=\frac{1}{0.0167\text{m}}$.
5. Convert $f$ to metres: $f=0.025\text{m}$. Then $\frac{1}{f}=40\text{D}$.
6. Using $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$, we have $40=60+\frac{1}{u}$, so $\frac{1}{u}=40-60=-20\text{D}$.
7. Thus $u=-\frac{1}{20}\text{m}=-0.05\text{m}=-5.0\text{cm}$. The negative sign indicates the object is on the same side as the incoming light (a real object).
8. Therefore, the least distance of distinct vision is $5.0\text{cm}$.

Concept tested: Accommodation and near point of the eye

Published 6 September 2026 · How this solution was prepared and checked