Question

A diverging lens, focal length f₁ = 20 cm, is separated by 5 cm from a converging mirror, focal length f₂ = 10 cm. Where an object should be placed so that a real image is formed at the object itself?  

Answer 1

: Let the object be placed at a distance x to the left of the lens. 

From the lens equation, 

(1/v) = (1/-20) + (1/-x) v = - [20x/(x+20)] 

A virtual image is formed due to first refraction at the lens. This image is an object for reflection from the concave mirror. Object distance is – 

[5+ {20x/ (20+x)}] = - [(25x +100)/(x+20)] 

From mirror equation, 

(1/v’) + [-(x+20)/(25x+100)] 

= (1/-10) (1/v’) 

= (-1/10) +[(x+20)/(25x+100)]

= [(10x +200-25x-100)/250(x+4)] 

v’ = -[50(x+4)/(3x-20)] 

This image is formed to the left of the mirror. Object distance for second refraction 

through concave lens, u’’ =- [5-{50(x+4)/ (3x-20)}] 

we assumed that second image lies between lens and mirror. The final image is produced at the object itself; hence v’’ = +x 

From lens equation, (1/x) – 1/[5-{(50x+4)/(3x-20)}]= 1/-20 on solving for x, we get 

25x² -1400x -6000 = 0 

x² -56x -240 = 0 

(x-60) (x+4) = 0 

Hence x = 60 cm 

The object must be placed at 60 cm to the left of the diverging lens.