Question

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?

Answer 1

Object distance=u=-8cm

Focal length f = 10 cm

Image distance v=?

`1/"v" - 1/"u" = 1/"f"`

`1/"v" - 1/-8 = 1/10`

`1/"v" = 1/10 - 1/8`

`1/"v" = (4 - 5)/40`

`1/"v" = 1/-40`

v = - 40 cm

As the object is placed between the focus and optical center of the lens, the image formed is virtual and erect.

Magnification (m) = `v/u = (-0)/-8`

m = +5.0

The +ve sign shows that the image formed is erect.