An object `3.0` cm high is placed perpendicular to the principal axis of a concave lens of focal length `7*5` cm. The image is formed at a distance of `5.0` cm from the lens. Calculate (i) distance at which object is placed, and (ii) size and nature of image formed.

Explanation

Here, `h_(1)=3.0 cm,f= - 7.5 cm, v= -5.0 cm,v=?, h_(2)=?`
From `(1)/(f)=(1)/(v)-1/u ,1/u=(1)/(v) - (1)/(f) =1/-5 - (1)/(-7.5) =(-3+2)/(15)` or `u = -15 cm`
i.e., object is at `15 cm` from the concave lens.
From `(h_(2))/(h_(1))=v/u, h_(2) =v/u xx h_(1)=(-5.0)/(-15.0)xx 3.0 =1cm`
The image is virtual and erect, and its size is 1cm.