Question

A normal eye has retina $2\text{cm}$ behind the eye-lens. What is the power of the eye-lens when the eye is most strained?

• A. $44\text{D}$
• B. $54\text{D}$
• C. $64\text{D}$
• D. $74\text{D}$

Answer 1

The correct option is B $54\text{D}$

Image always forms on retina.
$\therefore v=2\text{cm}$

Eye is most strained, when object is at normal distance of distinct vision.
$\therefore u=-25\text{cm}$

Now using,

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

$\Rightarrow \frac{1}{f}=\frac{1}{2}-\frac{1}{-25}=\frac{25+2}{50}$

$\therefore f=\frac{50}{27}\text{cm}$

$\therefore$ Power of the eye lens,

$P=\frac{1}{f}=\frac{27}{50}\times 100=54\text{D}$

Hence, $\left(B\right)$ is the correct answer.

Key Concept: When calculating focal-length or power of the eye lens for a normal eye. Image distance is equal to distance of retina from eye-lens.