A diverging lens of focal length 12 cm and a converging mirror of focal length 10 cm are placed coaxially at a separation of 5 cm. Where should an object be placed so that a real image is formed at the object itself ?

Answer

A, C

Explanation

Let the object be placed at a distance x 20m the lens farther away from the mirrror. For the convace lens `(^(st) refraction)` u = x, f= -20 cm From lens formula
`(1)/(upsilon) - (1)/(upsilon) = (1)/(f)`
`rArr (1)/(upsilon) = (1)/((-20)) + (1)/((-x))`
`rArr upsilon = -((20x)/(x + 20))`
So, the virtual image due to the first refraction lies same side as that of object. (A'B') This image becomes the object for the convace mirror,
`u = -(5+ (20x)/(x + 20))`
`=- (25x + 100)/(x +20))`
f = -10cm
From mirror equation,
`(1)/(upsilon) - (1)/(u) = (1)/(f)`
`rArr (1)/(upsilon) = (1)/(-10) + (x +20)/(25x + 100)`
`=(10x + 200- 25x - 100)/(250(x +4))`
`rArr upsilon = (250(x+4))/(100- 15x)`
`=-(250(x+4))/(15x-100)`
`=-(50(x+4))/((3x - 20))`
So this image is formed towards left of the mirror. Again for second refraction in concanve lens,
`u= -[(5-50(x +4))/(3x - 20)]` (assuming that)
image of mirror is formed between the lens and mirror 3x - 20, `upsilon +x` (since, the final image is produced on the object (A''B'') using lens formula,
`(1)/(upsilon) - (1)/(u) = (1)/(f)`
`rArr (1)/(x) + (1)([5-50(x xx 4)])/(3x - 20) = (1)/(-20)`
`rArr 25x^2 - 1400 x - 6000=0`
`rArr x^2 - 56x - 240 = 0`
`rArr (x- 60) (x +4) = 0`
So, x= 60 m The object should be placed at a distance 60 cm from the lens farther away from the mirror so that the final image is formed on itself.