Question

A person is having hypermetropic eye whose near point is at $125cm$. What is the power of the spectacles required for that person?

• A. $+1D$
• B. $-1D$
• C. $+3D$
• D. $-3D$

Answer 1

The correct option is C

$+3D$

Step 1. Given data:

Near point of the person's defected eye = $125cm$.

We know, near point for normal eye = $25cm$

Step 2. Formula used:

We know the lens formula is given by:

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

where $v=$image distance from the lens, $u=$object distance from the lens, $f=$focal length,

Step 3. Calculate the Power

We know that the near point of a defective eye is $1m$ and that of a normal eye is $25cm$.

So in this problem, they gave the data as same as written above. i.e.

near the point of defective eye $+$near point of normal eye = $125cm$

Thus,$u=-25cm$, and $v=-100cm$

Putting the given values in the formula:

$\frac{1}{-100}-\frac{1}{-25}=\frac{1}{f}$

$$\begin{gathered} \Rightarrow f=\frac{100}{3}=33.3cm \\ \Rightarrow f=0.33m \end{gathered}$$

Now Power = $\frac{1}{f}=\frac{1}{0.33}=+3D$

Hence, the correct option is option C.