An arrow of height 2.5 cm is situated vertically at a distance of 10 cm from a convex mirror of focal length 20 cm . Where will be image be formed? Determine its height. If the arrow is moved away from the mirror what will happen to its image?

Explanation

`"Here", u = -10 cm and f=20 cm`
`"We know", (1)/(v) + (1)/(u) = (1)/(f)`
`or, " " (1)/(v) + (1)/(-10)=(1)/(20)`
`or, " " (1)/(v) =(1)/(10)+(1)/(20)=(2+1)/(20)=(3)/(20)`
`or, " " v=(20)/(3) = 6.66 cm`
So, the image will be formed at a distance of 6.66 cm behind the convex mirror.
`"Magnification", m=(I)/(O)=-(v)/(u)`
`therefore " " I = -Oxx(v)/(u)`
`=-2.5 xx(20)/(3xx(-10))`
`=(5)/(3)=1.66 cm`
So, the height of the image = 1.66 cm.
If the arrow is moved away from the convex mirror the image will move towards the focus of the mirror and will be diminished in size.