Question

A person can read at the closest distance of $50cm$. In order for him to read at $30cm$. Identify the defect, nature, and power of the lens?

Answer 1

Step 1: Type of defect and type of lens used for correction

1. Myopia or nearsightedness is the defect here, it occurs when the eye losses its ability to focus on far-away objects.
2. Objects near the eye are visible to patients with this defect
3. The type of lens to be used here is a diverging lens to correct the defect

Step 2: Given

$u=$ Object distance from the optic center

$v=$ Image distance from the optic center

$f=$ the focal length of the lens

$$\begin{gathered} u=-30cm \\ v=-50cm \end{gathered}$$

Step 3: Formula used

from the lens formula,

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

Step 4: Calculation

$$\begin{gathered} \Rightarrow \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \Rightarrow \frac{1}{f}=\frac{1}{-50}+\frac{1}{30} \\ \Rightarrow f=-75cm \end{gathered}$$

Power$=\frac{100}{-75}=-1.33D$

Hence, the defect is myopia, the lens used is diverging and power is $-1.33D$.