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.

Given,

f = -10 cm

u = -20 cm

v = ?

Lens formula is —

$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \\[0.5em]$

Substituting the values in the formula, we get,

$\dfrac{1}{v} - \dfrac{1}{-20} = \dfrac{1}{-10} \\[0.5em] \dfrac{1}{v} + \dfrac{1}{20} = - \dfrac{1}{10} \\[0.5em] \dfrac{1}{v} = -\dfrac{1}{10} - \dfrac{1}{20} \\[0.5em] \dfrac{1}{v} = \dfrac{-2-1}{20} \\[0.5em] \dfrac{1}{v} = -\dfrac{3}{20} \\[0.5em] \Rightarrow v = -\dfrac{20}{3} \\[0.5em] \Rightarrow v = -6.67$

Therefore, the image in formed 6.67 cm in front of the lens.

As we know,

the formula for magnification of a lens is —

$m = \dfrac{v}{u} = \dfrac{h$i}{ho} \\[0.5em] Given,

u = -20 cm

v = -6.67 cm

height of object (h_o) = 5 cm

height of image (h_i) = ?

Substituting the values in the formula, we get,

$\dfrac{-6.67}{-20} = \dfrac{\text{h}$i}{5} \\[0.5em] \text{h}i = 5 \times \dfrac{-6.67}{-20} \\[0.5em] \text{h}_i = 1.67 \text{ cm} \\[0.5em]

Hence, the image is formed 6.67 cm in front of the lens, it is of size 1.67 cm and it is virtual (as the magnification is positive), erect and diminished.