Question

2 cm long pin is placed perpendicular to the principal axis of a lens of focal length 15 cm at a distance of 25 centimetres from the lens find the position of the image and its size. Why do we take the focal length as positive?

Answer 1

w = 2 cm
f = 15 cm
u = 25 cm
Now since the object is beyonf f of the lens, so the image formed will be real and on the right side of the lens. Whenever the image forms on the right side, the focal length is taken as positive according to the sign conventions.
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$=\frac{1}{15}=\frac{1}{v}+\frac{1}{25}$
$=\frac{1}{v}=\frac{5-3}{75}$
$=v=\frac{75}{2}cm$
$=37.5cm$
Also, magnification $=\frac{-v}{u}$
$=\frac{-37.5}{-25}=1.5$
Hence the image forward in 37.5 cm from less and magnification = 1.5 cm