Q52E - University Physics with Modern Physics

Question

Contact Lenses. Contact lenses are placed right on the eyeball, so the distance from the eye to an object (or image) is the same as the distance from the lens to that object (or image). A certain person can see distant objects well, but his near point is 45.0 cm from his eyes instead of the usual 25.0 cm . (a) Is this person nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct his vision? (c) If the correcting lenses will be contact lenses, what focal length lens is needed and what is its power in diopters?

Step-by-Step Solution

Answer

a) The person is farsighted.

b) Correction lens are converging lens.

v) The farsightedness can be corrected by using a converging lens of focal length +56.25cm and power +1.777 diopters.

Step 1: Basic definition

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.

Step 2: Focal length and Power of the lens

Object distance, u = + 25.0 cm (Near point of the eyes with lens)

Image distance, v = - 45.0 cm (Near point of the eyes without lens)

 

By using lens formula

 

$$\begin{gathered} \Rightarrow \frac{1}{+25}+\frac{1}{-45}=\frac{1}{f} \\ \Rightarrow \frac{45-25}{25*25}=\frac{1}{f} \\ \Rightarrow f=56.25cm \end{gathered}$$

 

Power of the lens is given by

 

$$\begin{gathered} \Rightarrow Power,P=\frac{1}{Focallength,f} \\ \Rightarrow P=\frac{1}{f}=\frac{100}{56.25}cm \\ =1.777 diopters \end{gathered}$$

 

Therefore, the farsightedness can be corrected by using a converging lens of focal length 

 ( 1.1 )                                                                    

+56.25 cm and power +1.777 diopters.