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?
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Since, the retina is 2cm behind the eye-lensv = 2cm
$\text{v}=2\text{cm}=0.02\text{m}$
$\Rightarrow\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{0.02}-\frac{1}{\infty}=50\text{D}$
So, in this condition power of the eye-lens is 50D
$\text{v}=+2\text{cm}=+0.02\text{m}$
$\Rightarrow\frac{1}{\text{u}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{0.02}-\frac{1}{-0.25}=50+4=54\text{D}$
In this condition power of the eye lens is 54D.
- When the eye-lens is fully relaxed
$\text{v}=2\text{cm}=0.02\text{m}$
$\Rightarrow\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{0.02}-\frac{1}{\infty}=50\text{D}$
So, in this condition power of the eye-lens is 50D
- When the eye-lens is most strained,
$\text{v}=+2\text{cm}=+0.02\text{m}$
$\Rightarrow\frac{1}{\text{u}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{0.02}-\frac{1}{-0.25}=50+4=54\text{D}$
In this condition power of the eye lens is 54D.
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