A farsighted person is unable to see objects placed closer than 50 cm . Determine the power of the lens required to see objects located at 20 cm .

Question

Quantrex Academy · Accepted Answer

Correct answer: B. +3 D

For a farsighted person, the lens must form a virtual image of the object at the person's near point.

Given object distance,

u = -20 \text{ cm}

Image distance,

v = -50 \text{ cm}

Using the lens formula:

\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}

\dfrac{1}{f} = \dfrac{1}{-50} - \dfrac{1}{-20}

\dfrac{1}{f} = -\dfrac{1}{50} + \dfrac{1}{20}

\dfrac{1}{f} = \dfrac{-2 + 5}{100} = \dfrac{3}{100} \text{ cm}^{-1}

The focal length of the lens is

f = \dfrac{100}{3} \text{ cm} = \dfrac{1}{3} \text{ m}

The power of the lens is given by

P = \dfrac{1}{f(\text{in m})}

P = +3 \text{ D}

A farsighted person is unable to see objects placed closer than 50...