A 15-cm-focal-length converging lens is 19 cm to the right of a 6.0-cm-focal-length converging lens. A 1.0-cm-tall object is distance L to the left of the 6.0-cm-focal-length lens. Required:
For what value of L is the final image of this two-lens system halfway between the two lenses?

L = 11.014 cm

Explanation:

Halfway between the two lenses is 19/2 = 9.5 cm.

Thus, this means virtually with respect to lens, the final image is at -9.5 cm

Thus, from here, we will work this out backwards.

Let's first solve for the initial position of the object for the second lens;

(1/S2) + (1/s'2) = (1/f2)

Where s'2 is the real image.

F2 is focal length

Thus;

(1/s'2) = (1/f2) - (1/s2)

(1/s'2) = (1/15) - (1/-9.5)

(1/s'2) = 0.1719

s'2 = 5.82 cm

The object for the second lens is located at 5.82 cm in front of the second lens.

Now, The object for the second lens and the image for the first lens will be the same.

This means the distance of the image from the first lens is at; 19 - 5.82 = 13.18 cm.

Now let's solve for the object distance of the first lens which will be denoted by L.

1/L = (1/f1) + (1/s'1)

Where f1 = 6 cm

1/L = (1/6) - (1/13.18)

1/L = 0.090794

L = 1/0.090794

L = 11.014 cm

divide the right side by the left side

0.024÷1.2=0.02 for example