An object 5 cm in length is held 25 cm away a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and nature of the image formed.

Explanation

`{:("Concave lens",,),(f=+10 cm,h=+5cm,h.=?),(,u=-25 cm,v=?),(,,"Nature of image = Erect"):}`
Lens formula `(l)/(v)-(l)/(u)=(l)/(f)`
`(l)/(v)-(l)/((-25))=(l)/(10), (l)/(v)-(l)/(10)-(l)/(25), (l)/(v)=(15)/(250)`
`v=+16.67 cm`
The image is formed at a distance of 16.67 cm on the right side of the convex lens (+ sign signifies this)
Magnification `m=(v)/(u)=(h.)/(h)`
`m=((16.67))/((-25))=(h)/(5), -0.67=(h.)/(5), " " h.=-3.35 cm`
The image formed (h.) has a height of 3.35 cm which is less than object height (h). The image formed is diminished. h. has a-ve sign which indicates that the image formed is real and inverted.