Question

A person uses spectacles of 'number' 2 for reading. Determine the range of magnifying power (angular magnification) possible. It is a concavo-convex lens (n = 10.5) having a curvature of one of its surfaces to be 10 cm. Estimate that of the other.

Answer 1

The magnifying power is the ratio of the least distance of distinct vision to the focal length of the lens and is maximum for close objects and minimum at infinity. The radius of curvature is related to the focal length of the lens by the lens-maker formula.

Formula used: Magnifying power at infinity $= \dfrac{D}{f}$ where $D$ is the least distance of distinct vision and $f$ is the focal length of the lens
Magnifying power at the nearest point $= 1 + \dfrac{D}{f}$
$\Rightarrow \dfrac{1}{f} = (n - 1)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ where $n$ is the refractive index of the lens, and ${R_1}$ ${R_2}$ are the radius of curvature of the two surfaces of the lens

Complete step by step answer:
For spectacles that have a ‘number’ 2 for reading, the focal length $f$ can be calculated as
$\Rightarrow f = \dfrac{1}{2} = 0.5\,m = 50\,cm$
The range of the magnifying power can be defined using the ratio of the least distance of distinct vision to the focal length of the lens.
The lowest magnification power will be at infinite distance and can be calculated as
$\Rightarrow {M_{min}} = \dfrac{D}{f}$
$\Rightarrow \dfrac{{25}}{{50}} = 0.5$
The highest magnification will be at the point of least distance of distinct vision and is calculated as
$\Rightarrow {M_{max}} = 1 + \dfrac{D}{f}$
$\Rightarrow 1 + 0.5 = 1.5$
To determine the radius of curvature of the surface, we use the lens formula
$\Rightarrow \dfrac{1}{f} = (n - 1)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Substituting the value of the refractive index $n = 10.5$ , ${R_1} = - 10$ (negative because the first surface is convex) we get
$\Rightarrow \dfrac{1}{{50}} = (10.5 - 1)\left( {\dfrac{1}{{ - 10}} - \dfrac{1}{{{R_2}}}} \right)$
$\Rightarrow \dfrac{1}{{475}} = \left( {\dfrac{1}{{ - 10}} - \dfrac{1}{{{R_2}}}} \right)$
Shifting the terms, we get
$\Rightarrow \dfrac{1}{{{R_2}}} = \left( {\dfrac{1}{{ - 10}} - \dfrac{1}{{475}}} \right)$
$\Rightarrow {R_2} = - 10.21\,cm$
Hence the radius of curvature of the second convex surface will be ${R_2} = - 10.21\,cm$ and the range of the magnifying power is from $0.5$ to $1.5$ .

Note:
The number of a pair of spectacles is calculated as the inverse of the focal length of the lens and we should remember this convention to find the focal length of the lens. While using the lens maker formula, we must be careful in taking into account the signs of the radius of curvature depending on whether the surface is concave or convex.