Question

A person with a myopic eye cannot see objects beyond $1.2\mathrm{m}$ distinctly. What should be the type of corrective lens used to restore proper vision?

Answer 1

Step 1. Formula to find the power of lens

In a normal eye, the distant point is infinity. The lens used should be such that an object at infinity, forms a virtual image at $1.2\mathrm{m}$.

Formula to find the power of lens is,

$\mathrm{P}=\frac{1}{\mathrm{f}}$

Here, $\mathrm{P}$ is the power of the lens and $\mathrm{f}$ is the focal length.

We know that the relation,

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

Here, $\mathrm{v}$ is the image distance and $\mathrm{u}$ is the object distance.

Step 2. Calculate the power of the corrective lens used to restore proper vision.

It is given that, a person can't see objects beyond $1.2\mathrm{m}$ directly.

So, in order to view the distances beyond $1.2\mathrm{m}$, from the given,

$\mathrm{v}=-1.2\mathrm{m}$

$\mathrm{u}=-\infty$

$\mathrm{f}=?$

Put the values in the above relation, we get,

$\frac{1}{\mathrm{f}}=\frac{1}{-1.2}-\frac{1}{\left(-\infty \right)}$

$\frac{1}{\mathrm{f}}=-\frac{1}{1.2}$

$\mathrm{f}=-1.2m$

Put the value of focal length in the power of lens formula, we get,

$\mathrm{P}=\frac{1}{-1.2}$

$=\frac{-5}{6}$

$=-0.83D$

So, the person suffers from Myopia, the corrective lens will be concave or diverging lens.

Hence, the power is $-0.833\mathrm{D}$ should be used to restore proper vision.