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Step-by-Step Solution
Step 1: Identify the Lenses and Their Parameters
• Diverging lens (concave lens) with focal length magnitude: $25\,\text{cm}$.
Because it is a diverging lens, its focal length is taken as $f_D = -25\,\text{cm}$.
• Converging lens (convex lens) with focal length magnitude: $20\,\text{cm}$.
Because it is a converging lens, its focal length is taken as $f_C = +20\,\text{cm}$.
• The distance between the two lenses is $d = 15\,\text{cm}$.
• A parallel beam of light is incident on the diverging lens first.
Step 2: Image Formation by the Diverging Lens
When a parallel beam of light is incident on a diverging lens, it forms an image at its focal point on the same side as the incident beam. Thus, for the diverging lens:
• Object distance $u_D = \infty$.
• Focal length $f_D = -25\,\text{cm}$.
• The image formed by this lens is at $v_D = f_D = -25\,\text{cm}$.
The negative sign indicates that the image is formed on the same side as the incident light (virtual image for the diverging lens).
Step 3: Treating the Image from the Diverging Lens as an Object for the Converging Lens
The image formed by the diverging lens now acts as the virtual object for the converging lens. The distance of this object from the converging lens is:
$u_C = v_D - d = (-25\,\text{cm}) - 15\,\text{cm} = -40\,\text{cm}.$
(The negative sign indicates that this object is on the same side as the incoming light for the converging lens.)
Step 4: Apply the Lens Formula to the Converging Lens
For the converging lens of focal length $f_C = +20\,\text{cm}$, using the lens formula:
$\displaystyle \frac{1}{f_C} = \frac{1}{v_C} - \frac{1}{u_C},$
substitute $f_C = +20\,\text{cm}$, $u_C = -40\,\text{cm}$:
$\displaystyle \frac{1}{20} = \frac{1}{v_C} - \left(-\frac{1}{40}\right) = \frac{1}{v_C} + \frac{1}{40}.$
Rearrange to solve for $v_C$:
$\displaystyle \frac{1}{v_C} = \frac{1}{20} - \frac{1}{40} = \frac{2 - 1}{40} = \frac{1}{40}.$
Hence,
$\displaystyle v_C = 40\,\text{cm}.$
This positive value of $v_C$ implies that the final image is formed on the opposite side of the converging lens relative to its object, indicating a real image.
Step 5: Conclusion
The final image is formed $40\,\text{cm}$ away from the converging lens on its right side (for real image). Hence, the correct option is:
Real and at a distance of 40 cm from the convergent lens.
