Find the position and magnification of the image of an object placed at distance of 8.0 cm in front of a convex lens of focal length 10.0 cm. Is the image erect or inverted?

i) As the image is formed on the other side of the lens, so the image is real. Hence, the lens is convex.

As we know, the lens formula is —

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

Given,

f = 10 cm

u = - 8 cm

Substituting the values in the formula, we get,

$\dfrac{1}{v} – \dfrac{1}{-8} = \dfrac{1}{10} \\[0.5em] \dfrac{1}{v} + \dfrac{1}{8} = \dfrac{1}{10} \\[0.5em] \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{8} \\[0.5em] \dfrac{1}{v} = \dfrac{4-5}{40} \\[0.5em] \dfrac{1}{v} = -\dfrac{1}{40} \\[0.5em] \Rightarrow v = - 40 cm \\[0.5em]$

Therefore, the image is formed 40 cm in front of the lens.

ii) As we know,

the formula for magnification of a lens is —

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

u = –8 cm

v = -40 cm

Substituting the values in the formula, we get,

$m = \dfrac{-40}{-8} \\[0.5em] m = 5 \\[0.5em]$

Therefore, the magnification is +5.0

As the magnification is positive so the image is erect.