(b)Determine the magnification in the following situation.
(c) Find if the magnifying power is equal to magnification.
Answer –
We have –
(a) We get the maximum possible magnification when the image is formed at the near point
So, Image distance is v = −d = −25 cm
Focal length is f = 10 cm
Object distance = u
Using the lens formula, we have:
$\frac{1}{f}$ = $\frac{1}{v}$ – $\frac{1}{u}$
$\frac{1}{u}$ = $\frac{1}{v}$ – $\frac{1}{f}$
= $\frac{1}{-25}$ – $\frac{1}{10}$ = -$\frac{7}{50}$
Therefore, u = $\frac{-50}{7}$
= – 7.14 cm
Hence, the lens should be kept at a distance of 7.14 cm in order to view the squares distinctly.
(b) Magnification = $\left | \frac{v}{u} \right |$
= $\frac{25}{\frac{50}{7}}$ = 3.5
(c) Magnifying power = $\frac{d}{u}$
= $\frac{25}{\frac{50}{7}}$
= 3.5
Since the image is formed at the near point (25 cm), the magnifying power is equal to the magnitude of magnification.