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?

Explanation

The minimum distance between object and image should be 4f to get real image on wall.
`therefore u + v = 4f`
`u = v= 3m`
`therefore 3 = 4 f`
`therefore f = 3/4 = 0.75` m
`rArr` Only for information :
`rArr` Suppose the object distance is u,
`therefore 1/ v - 1/u = 1/f`

`rArr` For convex lens
u is negative so, `1/v + 1/u = 1/f`
but u + v = d (is give)
`therefore v = d- u`
`therefore (1)/(d-u) + 1/u = 1/f`
`therefore (u + d-u)/(u(d-u)) = 1/f`
`therefore u^2 - ud + fd =0`
`rArr` This is the quadratic equation for variabel u. It.s roots are as given below :
`u = (d pm sqrt(d^2 -4 fd))/(2)`
But u = v
`therefore v = (d pm sqrt(d^2 - 4df))/(2)`
`rArr` For real image v is positive Hence, `d^2 - 4df ge 0 or d^2 ge 4 df `
`therefore d gt 4f`
Hence, to obtain the image , the miniumum value of d should be 4 f.
`therefore d = 4f and u + v = d `