1. Given data

• Focal length of lens $L_1$: $f_1 = +24\,\text{cm}$
• Focal length of lens $L_2$: $f_2 = +9\,\text{cm}$
• Separation between lenses: $10\,\text{cm}$
• Object distance from $L_1$: $u_1 = -6\,\text{cm}$

We use the lens formula:

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

with Cartesian sign convention.

2. Image formed by first lens $L_1$

For $L_1$:

$\frac{1}{24} = \frac{1}{v_1} - \frac{1}{(-6)}$

$\frac{1}{24} = \frac{1}{v_1} + \frac{1}{6}$

$\frac{1}{v_1} = \frac{1}{24} - \frac{1}{6} = \frac{1-4}{24} = -\frac{3}{24} = -\frac{1}{8}$

So,

$v_1 = -8\,\text{cm}$

This means the first lens forms a virtual image $8\,\text{cm}$ to the left of $L_1$.

3. Object distance for second lens $L_2$

The image formed by $L_1$ acts as the object for $L_2$.

• That image is $8\,\text{cm}$ to the left of $L_1$.
• Since $L_2$ is $10\,\text{cm}$ to the right of $L_1$, this point is

$10 + 8 = 18\,\text{cm}$

to the left of $L_2$.

Hence for $L_2$,

$u_2 = -18\,\text{cm}$

4. Final image formed by second lens $L_2$

Using lens formula for $L_2$:

$\frac{1}{9} = \frac{1}{v_2} - \frac{1}{(-18)}$

$\frac{1}{9} = \frac{1}{v_2} + \frac{1}{18}$

$\frac{1}{v_2} = \frac{1}{9} - \frac{1}{18} = \frac{2-1}{18} = \frac{1}{18}$

Therefore,

$v_2 = 18\,\text{cm}$

So the final image is formed $18\,\text{cm}$ to the right of $L_2$.

5. Distance between original object and final image

Take $L_1$ at $x=0$.

• Object is $6\,\text{cm}$ to the left of $L_1$: $x_O = -6\,\text{cm}$
• $L_2$ is at $x=10\,\text{cm}$
• Final image is $18\,\text{cm}$ to the right of $L_2$: $x_I = 10 + 18 = 28\,\text{cm}$

Hence distance between object and image is

$|x_I - x_O| = |28 - (-6)| = 34\,\text{cm}$

6. Final answer

$\boxed{34}$

The derived answer matches the stored correct answer.