Consider the situation described in the previous problem. Where should a point source be placed on the principal axis so that the two images form at the same place ?

Answer

A, C

Explanation

Let the source be placed at a distance 'x' from the lens as shown, so that image formed by both conicide. For the lense. `(1)/(upsilon) - (1)/(u) = (1)/(15)`
`rArr upsilon = (15x)/(x -15) ….. (i)`
For the mirror,
u = -(50-x)
f = -10cm
`So, (1)/(Vm) +(1)/((50-x)) = -(1)/(10)
`rArr = (1)/(Vm) = (1)/(50-x) - (1/(10)`
`=(10-50+x)/(10(50-x))`
`=(x-40)/(10(50-x))`
So, `V_m = (10(50 -x))/(x-40)....(ii)`
Since the lens and mirror are 50 cm apart,
`upsilon- V_m = 50`
`rArr (15x)/(x-15) - (10(50-x))/((x -40)) =50`
`rArr (3x^2 -120x) - (100x - 2x^2 -1500 +30x)`
`=10 (x^2 - 55 x+600)`
`rArr 5x^2 - 250x -1500 = 10x^2 - 550x + 6000`
`rArr 5x^2 - 300 x +4500 = 0`
`rArr x^2 -60x+900 = 0`
`rArr = (x-30)^2 = 0`
x =30 cm. so, the source should be placed 30 cm from the lens.