Question

What should be the distance between the object in Exercise 9.23 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2 . Would you be able to see the squares distinctly with your eyes very close to the magnifier? [Note: Exercises 9.22 to 9.24 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]

Answer 1

Area of the virtual image of each square, A= 6.25mm²
Area of each square, A°= 1mm²
Hence, the linear magnification of the object can be calculated as: m = \(\sqrt{\frac{A }{ A°}} = \sqrt{\frac{6.25 }{ 1}}= 2.5\)
But \(m = \frac{Image \space distance (v) }{ Object \space distance (u)}\)
∴ \(v = mu = 2.5u...(\)\(1)\)
Focal length of the magnifying glass, f = 10cm
According, to the lens formula, we have the relation: \(\frac{1 }{ƒ}= \frac{1}{v} - \frac{1 }{ u}\)
\(\frac{1}{10}= \frac{1 }{ 2.5u} - \frac{1 }{ u}= \frac{1 }{u} (\frac{1 }{ 2.5} -\frac{1 }{ 1}) = \frac{1}{u} (\frac{1-2.5 }{ 2.5})\)
\(∴ u = \frac{-1.5 \times 10 }{ 2.5} = -6cm\)
And v = 2.5u = 2.5 × 6 = -15cm
The virtual image is formed at a distance of 15cm, which is less than the near point (i.e., 25cm) of a normal eye, it cannot be seen by the eyes distinctly.