Q97P - University Physics with Modern Physics

Question

In one form of cataract surgery the person’s natural lens, which has become cloudy, is replaced by an artificial lens. The refracting properties of the replacement lens can be chosen so that the person’s eye focuses on distant objects. But there is no accommodation, and glasses or contact lenses are needed for close vision. What is the power, in diopters, of the corrective contact lenses that will enable a person who has had such surgery to focus on the page of a book at a distance of 24 cm?

Step-by-Step Solution

Answer

The power of the eye that will enable a person to focus on the page of a book at a distance of 24cm will be, 4.17D

Step 1:Relationship between object and image distance.

Object’s distance is s and image distance is s’.

The relationship is $\frac{1}{s}+\frac{1}{s\text{'}}=\frac{1}{\mathit{f}}$ 

Step 2:Formula for calculating power.

The power is calculated by finding the reciprocal of the focal length.

$\mathit{P}=\frac{1}{\mathit{f}}$ 

Step 3: Calculation of power of the eye.

The focal length using the relationship,

$$\begin{gathered} \frac{1}{s}+\frac{1}{s\text{'}}=\frac{1}{\mathit{f}} \\ \frac{1}{24\mathbf{cm}}+\frac{1}{\infty }=\frac{1}{\mathit{f}} \\ \mathit{f}=24\mathbf{cm} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathit{f}=0.24\mathbf{m} \end{gathered}$$ $$\begin{gathered} \frac{1}{s}+\frac{1}{s\text{'}}=\frac{1}{\mathit{f}} \\ \frac{1}{24\mathbf{cm}}+\frac{1}{\infty }=\frac{1}{\mathit{f}} \\ f=24\mathbf{cm} \\ f=0.24\mathbf{m} \end{gathered}$$

Therefore the power in diopters is,

$$\begin{gathered} \mathit{P}=\frac{1}{\mathit{f}} \\ =\frac{1}{0.24} \\ =4.17\mathit{D} \end{gathered}$$