**(a)** At what distance should the lens be held from the figure in order to view the squares distinctly with the maximum possible magnifying power?

**(b)** What is the magnification in this case?

**(c)** Is the magnification equal to the magnifying power in this case? Explain.

#### Solution

**(a) **The maximum possible magnification is obtained when the image is formed at the near point (d = 25 cm).

Image distance, v = −d = −25 cm

Focal length, f = 10 cm

Object distance = u

According to the lens formula, we have:

`1/"f" = 1/"v" - 1/"u"`

`1/"u" = 1/"v" - 1/"f"`

= `1/(-25) - 1/10`

= `(-2-5)/50`

= `- 7/50`

∴ u = `50/7` = −7.14 cm

Hence, to view the squares distinctly, the lens should be kept 7.14 cm away from them.

(b) Magnification = `|"v"/"u"| = 25/(50/7)` = 3.5

**(c) **Magnifying power = `"d"/"u" = 25/(50/7)` = 3.5

Since the image is formed at the near point (25 cm), the magnifying power is equal to the magnitude of magnification.