Question

Reading glasses of what power are needed for a person whose near point is 120 cm, so that she can read a computer screen at 55 cm?Assume a lens-eye distance of 2.0 cm.

Answer 1

Concepts and reason

The concept required to solve the given problem is lens formula and power of lens.

Initially, find the focal length of lens using equation $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ . In the next step, find the power of lens using equation $P = \frac{1}{f}$ .

Fundamentals

The expression for the lens equation is given as follows:

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

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

The power of the lens is,

$P = \frac{1}{f}$

The object distance from the glass is,

$\begin{array}{c}\\u = 55{\rm{ cm}} - 2.0{\rm{ cm}}\\\\ = 53{\rm{ cm}}\\\end{array}$

The image distance from the lens is,

$\begin{array}{c}\\v = 120{\rm{ cm}} - 2.0{\rm{ cm}}\\\\ = 118{\rm{ cm}}\\\end{array}$

The screen will produce a virtual image at a distance of 120 cm from the eye. Thus, the image distance is negative.

Substitute 53 cm for u and -118 cm for v in equation $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ .

$\begin{array}{c}\\\frac{1}{f} = \frac{1}{{ - 118{\rm{ cm}}}} + \frac{1}{{53{\rm{ cm}}}}\\\\f = 96.21{\rm{ cm}}\\\end{array}$

Substitute 96.12 cm for f in equation $P = \frac{1}{f}$ .

$\begin{array}{c}\\P = \frac{1}{{96.12{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1.0{\rm{ cm}}}}} \right)}}\\\\ = 1.039\\\\ \approx 1.04{\rm{ D}}\\\end{array}$

Ans:

The power of lens is 1.04 D.