Answer to Question #153630 in Optics for Charlton Wisner

Question

a small object is placed 6cm away from a converging lens of focal length 10cm. find the nature of the image, position of the image and magnification of the image

Answer 1

Distance of the object(u) = -6cm

Let the distance of the image be V cm

In regards to the sign convention in optics, focal length(f) of a converging lens is taken to be positive. Therefore, in this question f=+10cm

By implementing the lens formula:

$1/f=1/v-1/u$

$1/10=1/v-(-1/6)$

$1/10=1/v+1/6$

$1/10-1/6=1/v$

$1/v=-1/15$

$v=-15cm$

The image distance has a negative sign because the image is formed on the same side as that of the object of the lens .

$m=v/u$

$m=2.5$

signifying that the image formed is 2.5 times larger than the object

The positive value of the magnification obtained indicates that the image formed is erect.

The image is formed between 2F and infinity. Consequently, the image formed is virtual.