Question

The near point and the far point of a child are at $10cm$ and $100cm$ respectively. If the retina is $2cm$ behind the eye lens, the range of the power of the eye lens is:

• A. $40D$ to $20D$
• B. $60D$ to $51D$
• C. $35D$ to $22D$
• D. $55D$ to $25D$

Answer 1

The correct option is B $60D$ to $51D$

We know that power of a lens, $p=\frac{100}{f}$ D, where $f$ is focal length expressed in $cm$.
We will apply the lens formula to calculate the focal length ($f$).

Lens formula: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
In the first case (when the object is placed at the near point and image is formed on the ratina),

$v=2cm$ and $u=-10cm$

$\frac{1}{2}-\frac{1}{-10}=\frac{1}{f}$

$\Rightarrow \frac{5}{10}+\frac{1}{10}=\frac{1}{f}$

$\Rightarrow \frac{6}{10}=\frac{1}{f}$

$\Rightarrow f=\frac{10}{6}=\frac{5}{3}$

Thus, power $P=\frac{100}{f}=\frac{100\times 3}{5}=+60D$
In the second case (when the object is placed at the far point and image is formed on the ratina),

$v=2cm$ and $u=-100cm$

$\frac{1}{2}-\frac{1}{-100}=\frac{1}{f}$

$\Rightarrow \frac{50}{100}+\frac{1}{100}=\frac{1}{f}$

$\Rightarrow \frac{51}{100}=\frac{1}{f}$

$\Rightarrow f=\frac{100}{51}$

Thus, power $P=\frac{100}{f}=\frac{100\times 51}{100}=+51D$

Therefore, the range of the power of the eye lens is $60D$ to $51D$