A point object O is placed on the principle axis of a convex lens of focal length 10 cm  at 12 cm from the lens.

 U = -12

f = -10 

1/v - 1/u = 1/f

1/v + 1/12 = 1/10

1/v = 1/60 

\(-\frac{1}{v^2} \)dv or \(\frac{1}{u^2} \)du

du = \(\frac{v^2}{u^2} \)du

\( \frac{v}{u} \)= same path

△u₁ = △u₂ = 1mm

△u₁ = △v₂ 

△v₂ = v_f - v_i 

x₂ = △v₂ + 1mm

x₂ = 2mm

\(\frac{x_1}{x_2}=\frac{1}{2} \)

Given u = -10

f = 12

\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

\(\frac{1}{v}-(\frac{1}{-10})=\frac{1}{12}\)

\(\frac{1}{v}=\frac{1}{12}-\frac{1}{10}\)

\(\frac{1}{v}=\frac{10-12}{120}\)

\(\frac{1}{v}=\frac{-2}{120}\)

\(\frac{1}{v}=\frac{-1}{60}\)

\(-\frac{1}{v^2}\) dv + \(\frac{1}{u^2}\) du

du = \(\frac{v^2}{u^2}\) du

\(\frac{v}{u}\) = same path

△u₁ = △u₂ = 1mm

△u₁ = △v₂

△v₂ = v_f - v_i

x₂ = △v₂ + 1mm

x₂ = 2mm

\(\frac{x_1}{x_2}=\frac{1}{2}\)