Answer:
The distance between the two lenses is 4 cm.
Explanation:
The focal length f of the converging lens = 6 cm
The object distance u from the converging lens = -24 cm
Let the image distance made by converging lens be v.
By lens equation,
1/f = 1/v - 1/u
1/6 = 1/v - 1/(-24)
v = 8 cm
In the figure, image forms at point A. This image is object for second diverging lens. Let the distance of diverging lens from this point be x and distance from the screen be t.
OA = 8 cm and AO' = t - x = 10 - 8 = 2 cm
The image distance for diverging lens is t and object distance is x. Focal length of diverging lens is = -12 cm.
By applying lens formula to diverging lens,
1/(-12) = 1/t - 1/x
But t = x+2
1/-12 = 1/(2+x) - 1/x
[tex] 24 = 2x + x^{2} [/tex]
[tex] x^{2} + 2x - 24 = 0 [/tex]
[tex] (x-4)(x-6) = 0 [/tex]
Now, x has two roots 4 and -6. But x can not be negative. So, x = 4 cm.
t = x + 2 = 6 cm
The distance of screen from converging lens is 10 cm and t = 6 cm.
So, the length between converging lens and diverging lens = 10 - 6 = 4 cm.