Question

A person struggles to read by holding a book at arm's length, a distance of 45 cm away (= near point). What power of reading glasses should be prescribed for him, assuming they will be placed2.0 cm from the eye and he wants to read at the normal near point of 25 cm?

* answer: Power (P) = 2.02 D

Answer 1

Concepts and reason

The required concepts to solve this question are long site, converging lens, and power of lens.

Use the concept of converging lens to finding the corrective lens. Use the lens maker formula to find the focal length. Finally use the concept of power of lens to find the power of corrective lens.

Fundamentals

Converging lens produces a real image by converting the parallel light rays into converging light rays.

The expression for the focal length is,

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

Here, f is the focal length, u is the object distance, and v is the image distance.

The expression for the power is,

$P = \frac{{100}}{{f{\rm{ cm}}}}{\rm{ D}}$

The object distance from the lens is,

$\begin{array}{c}\\u = 25{\rm{ cm - 2 cm}}\\\\{\rm{ = 23 cm }}\\\end{array}$

The image distance from the lens is,

$\begin{array}{c}\\v = 45{\rm{ cm - 2 cm}}\\\\{\rm{ = 43 cm }}\\\end{array}$

Substitute 23 cm for u and -43 cm for v in the expression $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ .

$\begin{array}{c}\\\frac{1}{f} = \frac{1}{{23{\rm{ cm}}}} + \frac{1}{{ - 43{\rm{ cm}}}}\\\\f = \frac{{\left( { - 43{\rm{ cm}}} \right)\left( {23{\rm{ cm}}} \right)}}{{\left( {23 - 43} \right){\rm{ cm}}}}\\\\f = \frac{{\left( {43{\rm{ cm}}} \right)\left( {23{\rm{ cm}}} \right)}}{{\left( {43 - 23} \right){\rm{ cm}}}}\\\\ = 49.45{\rm{ cm}}\\\end{array}$

Substitute 49.45 cm for f in the expression $P = \frac{{100}}{{f{\rm{ cm}}}}$ .

$\begin{array}{c}\\P = \frac{{100}}{{49.45{\rm{ cm}}}}{\rm{ D}}\\\\{\rm{ = 2}}{\rm{.02 D}}\\\end{array}$

Ans:

The power of the corrective lens (converging lens) is 2.02 D