Q53E - University Physics with Modern Physics

Question

Ordinary Glasses: Ordinary glasses are worn in front of the eye and usually 2.0 cm in front of the eyeball. Suppose that the person in previous exercise prefers ordinary glasses to contact lenses. What focal length lenses are needed to correct his vision, and what is their power in diopters?

Step-by-Step Solution

Answer

The farsightedness can be corrected by using a converging lenses of focal length +49.45 and power +2.022 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.

Step 2: Focal length and Power of the Lens

Object distance, u = (+25.0-2)cm  

                               =+23cm

 Image distance,  v = (+45.0+2)cm 

                                 = - 43 cm

By using lens formula,   

$$\begin{gathered} \Rightarrow \frac{1}{+23}+\frac{1}{-43}=\frac{1}{f} \\ \Rightarrow \frac{43-23}{43*23}=\frac{1}{f} \\ \Rightarrow f=+45.49 cm \end{gathered}$$

 

Power of the lens is given by

 $$\begin{gathered} \Rightarrow Power,P=\frac{1}{Focal length,f} \\ \Rightarrow P=\frac{1}{f}=\frac{100}{49.45}cm \\ =2.022 diopters \end{gathered}$$

Therefore, the farsightedness can be corrected by using a converging glasses of focal length +49.45 cm and power +2.022 diopters.