Question

An object is 5 cm in front of a convex lens of focal length 7.5 cm.What is the position and magnification of the image.​

Answer 1

u = -5 cm

f = 7.5 cm

$\sf \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\\\ \sf \frac{1}{7.5} = \frac{1}{v} - \frac{1}{( - 5)} \\\\ \sf \frac{1}{v} = \frac{1}{7.5} + \frac{1}{( - 5)} \\\\ \sf \frac{1}{v} = \frac{10}{75} - \frac{1}{5} \\\\ \sf \frac{1}{v} = \frac{10 - 15}{75} \\\\ \sf \frac{1}{v} = \frac{ - 5}{75} \\\\ \sf v = - 15 \: cm$

$\sf m = \frac{v}{u} \\\\ \sf m = \frac{ - 15}{5} \\\\ \sf m = - 3$