Question

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

Answer 1

Let,
`upsilon = +upsilon`
`therefore u=-(3-upsilon)`
`f_("max")=?`
Now, `(1)/(f)=(1)/(upsilon)-(1)/(u)=(1)/(upsilon)-(1)/(-(3-upsilon))`
`=(1)/(f)=(1)/(upsilon)+(1)/(3-upsilon) rArr (1)/(f)=(3-upsilon +upsilon)/((3-upsilon)upsilon)`
`3upsilon - upsilon^(2)=3f`
For f to be maximum `d(f)=0`
i.e. `d(3upsilon-upsilon^(2))=0`
`3-2upsilon=0`
`upsilon = 3//2=1.5m`
Hence, `u=-(3-1.5)`
`=-1.5m`
and `(1)/(f)=(1)/(v)-(1)/(u)=(1)/(1.5)-(1)/(-1.5)=(1+1)/(1.5)`
`=(2)/(1.5)=0.75m`