A `2.0 cm` tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position and size of the image. Also, find its magnification.

Explanation

Height of the object h = + 2.0 cm,
Focal length f = + 10 cm,
object-distance u = –15 cm,
Image-distance v =?
Height of the image h′ =?
Since `1/v-1/u=1/f`
or, `1/v=1/u+1/f`
`1/v=1/(-15)+1/10= -1/15+1/10`
1/v=(-2+3)/30=1/30`
or,`" "v= +30 cm`
The positive sign of v shows that the image is formed at a distance of 30 cm on the other side of the optical centre. The image is real and inverted.
Magnification `m=(h')/h=v/u`
or,`" "h'=h(v//u)`
Height of the image, `h′ = (2.0) (+30/–15) = – 4.0 cm`
Magnification m = v/u
or, `m=(+30cm)/(-15cm)= -2`
The negative signs of m and h′ show that the image is inverted and real. It is formed below the principal axis. Thus, a real, inverted image, 4 cm tall, is formed at a distance of 30 cm on the other side of the lens. The image is two times enlarged.