Ncert Exercises & Solutions

9.31:The virtual image of each square in the figure is to have an area of 6.25 mm². Find out, what should be the distance between the object in Exercise 9.30 and the magnifying glass? If the eyes are too close to the magnifier, would you be able to see the squares distinctly?

Hint: $m=\sqrt{\frac{A}{A_0}}$

Step 1: Find the image distance.
Given that:

Area of the virtual image of each square, A = 6.25 mm²

Area of each square, A₀ = 1 mm²

The linear magnification:

$m = \sqrt{\frac{A}{A_{0}}}$
$= \sqrt{\frac{6.25}{1}} = 2.5$
$We know, m = \frac{lmage distance (v)}{Object distance (u)}$
$∴ v = mu = 2.5 u$

Step 2: Find the object's distance.
The focal length of the lens, f = 10 cm

According to the lens formula,

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{10} = \frac{1}{2.5 u} - \frac{1}{u} = \frac{1}{u} (\frac{1}{2.5} - \frac{1}{1}) = \frac{1}{u} (\frac{1 - 2.5}{2.5})$
$∴ u = - \frac{1.5 \times 10}{2.5} = - 6 cm$
$And v = 2.5 u = 2.5 \times 6 = - 15 cm$

As the image of the squares is formed at a distance of 15 cm, which is less than the near point (i.e. 25 cm) of a normal eye. Hence, it cannot be seen by the eyes distinctly.

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