An object of height 5 cm is placed perpendicular to the principal axis of a concave lens of focal length 10 cm. If the distance of the object from the optical centre of the lens is 20 cm, determine the position, nature and size of the image formed using the lens formula.
CBSE OUTSIDE DELHI - SET 3 2015
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$h = + 5\text{ cm} \ \ \ \ \ f = -10 \text{ cm} \ \ \ \ \ u = -20 \text{ cm} \ \ \ \ \ v = ?$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\therefore \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{(-10)} + \frac{1}{(-20)} \ \ \ \ \ \ = \frac{-2-1}{20} = \frac{-3}{20}$
$\therefore v = -\frac{20}{3} \text{ cm}$
Nature – Virtual, erect
$h' = \frac{v}{u} \times h = \bigg(\frac{-20}{3} \times \frac{1}{-20}\times5\bigg) \text{ cm} = + \frac{5}{3} \text{ cm}$
$\therefore \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{(-10)} + \frac{1}{(-20)} \ \ \ \ \ \ = \frac{-2-1}{20} = \frac{-3}{20}$
$\therefore v = -\frac{20}{3} \text{ cm}$
Nature – Virtual, erect
$h' = \frac{v}{u} \times h = \bigg(\frac{-20}{3} \times \frac{1}{-20}\times5\bigg) \text{ cm} = + \frac{5}{3} \text{ cm}$
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