1. Given data

• Diverging lens first: focal length $f_1=-25\text{ cm}$
• Converging lens second: focal length $f_2=+20\text{ cm}$
• Distance between lenses $=15\text{ cm}$
• Incident light on first lens is parallel, so object for first lens is at infinity.

2. Image formed by the diverging lens

For an object at infinity, the image is formed at the focus of the lens.

Since the first lens is diverging, the image is virtual and formed on the same side as the object at a distance equal to its focal length:

$v_1=f_1=-25\text{ cm}$

So the first lens forms a virtual image $25\text{ cm}$ to its left.

3. This image acts as object for the converging lens

The second lens is $15\text{ cm}$ to the right of the first lens.

Hence the virtual image formed by the first lens is at a distance from the second lens:

$25+15=40\text{ cm}$

This point lies to the left of the converging lens, so for the second lens the object distance is

$u_2=-40\text{ cm}$

(using Cartesian sign convention).

4. Apply lens formula to the converging lens

Lens formula:

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

For the second lens:

$\frac{1}{20}=\frac{1}{v_2}-\frac{1}{(-40)}$

$\frac{1}{20}=\frac{1}{v_2}+\frac{1}{40}$

$\frac{1}{v_2}=\frac{1}{20}-\frac{1}{40}=\frac{1}{40}$

$v_2=40\text{ cm}$

Since $v_2$ is positive, the final image is real and forms $40\text{ cm}$ to the right of the converging lens.

5. Match with options

• A: real and $6\text{ cm}$ from converging lens $\rightarrow$ incorrect
• B: real and $40\text{ cm}$ from converging lens $\rightarrow$ correct
• C: virtual and $40\text{ cm}$ from converging lens $\rightarrow$ incorrect
• D: real and $40\text{ cm}$ from divergent lens $\rightarrow$ incorrect, because from divergent lens it would be $15+40=55\text{ cm}$

6. Final answer

The final image is real and at a distance of $40\text{ cm}$ from the converging lens.