Question

A normal eye has retina $2cm$ behind the eye-lens. What is the power of the eye-lens when the eye is fully relaxed & most strained?

• A. $50D,54D$
• B. $25D,54D$
• C. $50D,30D$
• D. $20D,45D$

Answer 1

Answer: (A)

$50D,54D$

For case: I) $v=\infty ,u=2cm=0.02m$
$⟹\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{0.02}-\frac{1}{\infty }=50D$
So, power of the lens is $50D$
For case II) Most strained
$u=-25cm,0.25m$
$v=+2cm=.02m$
$⟹\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{0.02}+\frac{1}{0.25}=50+4=54D$
Power of the lens is $54D$.