A converging lens has a focal length of `30 cm.` Rays from a `2.0 cm` high filament that pass through the lens from a virtual image at a distance of `50 cm` from the lens. Where is the filament located? What is the height of the image?

Answer

A, C

Explanation

`1/-50-1/u=1/(+30)`
Solving, we get `u=-18.75 cm`
`m=v/u=((-50))/((-18.75))=2.67`
`I=m(O)=2.67xx2`
`=5.33 cm`