A small object of height `0.5` cm is placed in front of a convex surface of glass `(mu-1.5)` of radius of curvature 10cm. Find the height of the image formed in glass.

Explanation

According to cartesian sign convention,
`u=-30cm, R=+10cm, mu_(1)=1,mu_(2)=1.5`
Applying the equation, we get ` (1.5)/(v)=(1)/(-30)=(1.5-1)/(+10)`
or `v=90cm` (real image)
Let `h_(1)` be the height of the image, then
`(h_(i))/(h)=(mu_(1)v)/(mu_(2)u)=((1)(90))/((1.5)(-30))=-2`
`rArrh_(i)=-2h_(0)(0.5)=-2(0.5)=-1cm`
The negative sign shows that the image is inverted.