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

If both images are formed at same place, for this one image should be real and another virtual. Let image of `O_1` is real.
Image of `O_1`: `u=-x,f=9cm`
`(1)/(v_1)-(1)/(-x)=(1)/(9)implies(1)/(v_1)=(1)/(9)-(1)/(x)`
Image of `O_2`: `u=-(24-x),f=9cm`
`(1)/(v_2)-(1)/(-(24-x))=(1)/(9)implies(1)/(v_2)=(1)/(9)-(1)/((24-x))`
For `O_1`. Direction of incident rays is from left to right hence `v_1` is positive
for `O_2`. Directioin of incident rays if from right to left hence `v_2` is negative. ltbegt `(1)/(v_1)=(1)/(v_2)`
`(1)/(9)-(1)/(x)=-((1)/(9)-(1)/(24-x))implies(2)/(9)=(1)/(24-x)+(1)/(x)`
`(2)/(9)=(24)/((24-x)x)impliesx^2-24x+108=0`
`(x-6)(x-18)=0`
`x=6cm` or `x=18cm`