The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall `3 m` away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose ?

Explanation

Let image distance, `v=+v`
`therefore" Object distance, u"=-(3-v)`,
Maximum possible focal length of lens, `f_("max")=?`
`"Now "(1)/(f)=(1)/(v)-(1)/(u)=(1)/(v)-(1)/(-(-3-v))`
`=(1)/(v)+(1)/(3-v)`
`"or "(1)/(f)=(3-v+v)/((3-v)v)`
`"or "3v-v^(2)=3f`
For f to be maximum `d(f)=0`
`"i.e. "d(3v-v^(2))=0`
`"i.e. "3-2v=0`
`v=(3)/(2)=1.5m`
Hence, `u=-(3-1.5)=-1.5m`
`"and "(1)/(f_("max"))=(1)/(v)-(1)/(u)=(1)/(1.5)-(1)/(-1.5)`
`=(1+1)/(1.5)`
`f_("max")=(1.5)/(2)=0.75m.`