A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision. (a) Is she nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct her vision? (c) What focal length contact lens is needed, and what is its power in diopters?

Thin Lens formula:

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

Here,

u = Object distance from lens

v = Image distance from lens

f = Focal length of the lens

Sign Convention:

1. Object Distance (u): (+) in front of lens; (-) in the back of lens
2. Image Distance (v): (-) in front of lens; (+) in the back of lens
3. Focal Length (f): (+) for convergent (convex) lens; (-) for divergent (concave) lens

Power of a lens is defined as the reciprocal of its focal length. Thus, converging or convex lens have positive powers and diverging or concave lens have negative powers.

When a person can see far objects clearly, but near objects are blurred then such condition is known as farsightedness and the person is known as farsighted person. To correct the vision of such a person, converging or convex lenses are used.

When a person can see near objects clearly, but far objects are blurred then such condition is known as Nearsightedness and the person is known as nearsighted person. To correct the vision of such a person, diverging or concave lenses are used.

Object distance, $u=\infty$(Far point of the eyes with lens)

Image distance, v = - 75.0 cm (Near point of the eyes without lens)

By using lens formula

$$\begin{gathered} \Rightarrow \frac{1}{\infty }+\frac{1}{-75}=\frac{1}{f} \\ \Rightarrow -\frac{1}{75}=\frac{1}{f} \\ \Rightarrow =-75cm \end{gathered}$$

Power of the lens is given by

$$\begin{gathered} \Rightarrow Power,P=\frac{1}{Foacllength,f} \\ \Rightarrow P=\frac{1}{f}=\frac{100}{-75}cm \\ =-1.333diopters \end{gathered}$$

Therefore, the Nearsightedness can be corrected by using a diverging lens of focal length -75 cm and power -1.333 diopters.