Question

A person needs a lens of power $- 5.5$ diopters for correcting his distant vision. For correcting his near vision, he needs a lens of power $+ 1.5$ diopter. What is the focal length of the lens required for correcting (i) distance vision, and (ii) near vision?

Answer 1

We will first write down the given information and then by using the formula, i.e., $p = \dfrac{1}{f}$, where $p$ is the power of the lens and $f$ is the focal length of the lens, we will find the required answers for both the parts.

Complete step-by-step solution:
We know that the power of a lens is a measure of the degree of convergence or divergence of the light rays falling on it. The power of a lens may also be defined as the reciprocal of its focal length. Therefore, $p = \dfrac{1}{f}$.
Now, by using this formula we will solve both the parts.
(i) Distance vision
Given that,
Power of the lens is $P = - 5.5D$
We know that, the formula for power is $p = \dfrac{1}{f}$
Hence, the focal length is given by

\Rightarrow f = \dfrac{1}{{ - 5.5}} \\\ \Rightarrow f = - \dfrac{{10}}{{55}} = - \dfrac{2}{{11}} = 0.181\text{ m} \\\ \therefore f = - 18.1{\text{ cm}} $$ Hence, the focal length of the lens required for correcting distance vision is $$ - 18.1\,\,{\text{cm}}$$ (ii) Near vision Given that power of the lens is $$P = 1.5{\text{D}}$$ We know that, the formula for power is $$p = \dfrac{1}{f}$$ Hence, the focal length is given by $$ \Rightarrow f = \dfrac{1}{P} \\\ \Rightarrow f = \dfrac{1}{{1.5}} \\\ \Rightarrow f = \dfrac{{10}}{{15}} = \dfrac{2}{3}= 0.667\text{ m} \\\ \therefore f = 66.7{\text{ cm}} $$ **Hence, the focal length of the lens required for correcting near vision is 66.7cm.** **Note:** The normal distance vision is 20/20 i.e., if we have 20/20 vision, we can see clearly an object at a distance from 20 feet. The near vision of an average person is 25 cm. Here diopter ‘$$D$$’ is the unit of power of a lens. Here we get the focal length in meters, we multiply this by 100 to convert in cm.