Optics 7 Question 2
2. An upright object is placed at a distance of $40 cm$ in front of a convergent lens of focal length $20 cm$. A convergent mirror of focal length $10 cm$ is placed at a distance of $60 cm$ on the other side of the lens. The position and size of the final image will be
(2019 Main, 8 April I)
(a) $20 cm$ from the convergent mirror, same size as the object
(b) $40 cm$ from the convergent mirror, same size as the object
(c) $40 cm$ from the convergent lens, twice the size of the object
(d) $20 cm$ from the convergent mirror,twice size of the object
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Answer:
Correct Answer: 2. (*)
Solution:
- In given system of lens and mirror, position of object $O$ in front of lens is at a distance $2 f$.
i.e. $u=2 f=40 cm$
So, image $\left(I _1\right)$ formed is real, inverted and at a distance, $v=2 f=2 \times 20=40 cm$, (behind lens) magnification,
$m _1=\frac{v}{u}=\frac{40}{40}=1$
Thus, size of image is same as that of object.
This image $\left(I _1\right)$ acts like a real object for mirror.
As object distance for mirror is $u=C=2 f=-20 cm$ where, $C=$ centre of curvature.
So, image $\left(I _2\right)$ formed by mirror is at $2 f$.
$\therefore$ For mirror $v=2 f=2(-10)=-20 cm$
Magnification, $m _2=-\frac{v}{u}=-\frac{(-20)}{(-20)}=-1$
Thus, image size is same as that of object.
The image $I _2$ formed by the mirror will act like an object for lens.
As the object is at $2 f$ distance from lens, so image $\left(I _3\right)$ will be formed at a distance $2 f$ or $40 cm$. Thus, magnification,
$$ m _3=\frac{v}{u}=\frac{40}{40}=1 $$
So, final magnification, $m=m _1 \times m _2 \times m _3=-1$
Hence, final image $\left(I _3\right)$ is real, inverted of same size as that of object and coinciding with object.
