The distance between two point sources of light is `24cm.` Find out where would you place a converging lens of focal length `9cm,` so that the images of both the sources are formed at the same point.

Explanation

Images of both the objects at the same location is possible only when one of the images is real and the other is virtual. Refer to the following ray diagram:

Here, in the ray diagram we have assumed real image for object `O_(1)` and virtual image for object `O_(2)` both at a distance y from the lens. Distance between `O_(1)` and lens is assumed to be x.
Applying lens formula for `O_(1)`.
`(1)/(v)-(1)/(u)=(1)/(f)`
`rArr" "(1)/(+y)-(1)/(-x)=(1)/(9)`
`rArr" "(1)/(y)=(1)/(9)-(1)/(x)`
`rArr" "y=(9x)/(x-9)" ...(i)"`
Applying lens formula for `O_(2)`.
`(1)/(v)-(1)/(u)=(1)/(f)`
`rArr" "(1)/(-y)-(1)/(-(24-x))=(1)/(9)`
`rArr" "(1)/(y)=(1)/(24-x))-(1)/(9)`
`rArr" "(1)/(y)=(9-24+x)/(9(24-x))`
`rArr" "y=(9(24-x))/(x-15)" ....(ii)"`
From equations (i) and (ii) we can write the following :
`(9(24-x))/(x-15)=(9x)/(x-9)`
`rArr" "((24-x))/(x-15)=(x)/(x-9)`
`rArr" "24x-216-x^(2)+9x=x^(2)-15x`
`rArr" "2x^(2)-48x+216=0`
`rArr" "x^(2)-24x+108=0`
`rArr" "x^(2)-18x-6x+108=0`
`rArr" "x(x-18)-6(x-18)=0`
`rArr" "(x-6)(x-18)=0`
`rArr" "x=6cm and 18 cm`
Hence lens is kept at a distance 6 cm from one and thus 18 cm from the other point source.