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Step-by-Step Solution
Step 1: Identify the Given Parameters
• A convergent lens (thin convex lens) of focal length
$f_{\text{lens}} = 20\,\text{cm}$.
• An object placed $40\,\text{cm}$ in front of the lens (i.e., object distance
$u_{\text{lens}} = -40\,\text{cm}$).
• A convergent mirror (concave mirror) of focal length
$f_{\text{mirror}} = 10\,\text{cm}$ placed $60\,\text{cm}$ behind the lens.
Because the object is placed at $u_{\text{lens}} = -40\,\text{cm}$ for the lens with
$f_{\text{lens}} = 20\,\text{cm}$, notice that $|u_{\text{lens}}| = 2 f_{\text{lens}}$,
which implies the object is placed at $2f$ for the lens.
Step 2: Determine Image Formed by the Lens
When an object is placed at $2f_{\text{lens}}$ from a convex lens, the image is formed
at $2f_{\text{lens}}$ on the other side of the lens. Hence,
$v_{\text{lens}} = 40\,\text{cm}$.
The magnification of the lens
$m_1 = \frac{v_{\text{lens}}}{u_{\text{lens}}}
= \frac{40}{40} = 1$.
Thus, the image ($I_1$) is real, inverted, and of the same size as the object.
Step 3: Image of Lens Acts as Object for the Mirror
This image ($I_1$) lies $60 - 40 = 20\,\text{cm}$ in front of the mirror
(because the mirror is placed $60\,\text{cm}$ from the lens, and $I_1$ is
$40\,\text{cm}$ from the lens on the same side). However, for sign convention
for the mirror, this distance is taken as negative (object in front of mirror).
Hence $u_{\text{mirror}} = -20\,\text{cm}$.
Step 4: Determine the Image Formed by the Mirror
Since the mirror has $f_{\text{mirror}} = 10\,\text{cm}$, note that
$|u_{\text{mirror}}| = 2 f_{\text{mirror}}$. A concave mirror with an object
at $2f$ forms the image at $2f$ itself.
Therefore, $v_{\text{mirror}} = -20\,\text{cm}$ (behind the mirror, on the
same side as the object, by sign convention for mirrors).
Magnification by the mirror
$m_2 = -\frac{v_{\text{mirror}}}{u_{\text{mirror}}}
= -\frac{-20}{-20} = -1$.
This image is also the same size as the object (but inverted relative to $I_1$).
Step 5: Mirror's Image Acts as Object for the Lens Again
The image formed by the mirror ($I_2$) now travels back through the lens.
Since it originates $20\,\text{cm}$ from the mirror, it reenters the lens
at a distance of $40\,\text{cm}$ from the lens (the same geometry in reverse).
Step 6: Final Image Position by the Lens
Once again, the lens sees an object at $2 f_{\text{lens}} = 40\,\text{cm}$.
So the final image will be formed at $2 f_{\text{lens}} = 40\,\text{cm}$ from the lens.
The magnification by this final pass through the lens
$m_3 = \frac{v_{\text{lens}}}{u_{\text{lens}}}
= \frac{40}{40} = 1$.
Step 7: Overall Magnification and Conclusion
The total magnification is
$m = m_1 \times m_2 \times m_3
= (1) \times (-1) \times (1) = -1$,
indicating the final image is inverted and has the same size as the original object.
Therefore, the final image forms $40\,\text{cm}$ from the lens and is of the same
size as the object (though inverted).
Final Answer
"40 cm from the convergent lens, same size as the object."
