The near and far points of a person are at `40 cm` and `250cm` respectively. Find the power of the lens he/she should use while reading at `25cm`. With this lens on the eye, what maximum distance is clearly visible?

Options

A. `2.5 D`
B. `5.0 D`
C. `1.5 D`
D. `3.5 D`

Answer

C

Explanation

In the eye of the power there is a defect for seeing the near objects as well as far off objects. He intends to read a book keeping it at a distance `25 cm` from the eye. The lens used should be such that the image of this object by lens shoud be formed at distance `40 cm` from eye lens, so that the final image formed by eye lens is very clearly seen.
Taking refraction from the lens used, we have
`u = -25 cm, v = -40 cm` ,
The lens formula is `(1)/(f) = -(1)/(u) + (1)/(v)`
or `(1)/(f) = (1)/(-25) + (1)/(-40) = (1)/(25) - (1)/(40) = (3)/(200)`
or `f = (200)/(3) cm = (2)/(3) m`
Power of lens `P = (1)/(f) = (1)/(2//3) = + 1.5 D`.