Optics 3 Question 18
18. The size of the image of an object, which is at infinity, as formed by a convex lens of focal length $30 cm$ is $2 cm$. If a concave lens of focal length $20 cm$ is placed between the convex lens and the image at a distance of $26 cm$ from the convex lens, calculate the new size of the image. (2003, 2M)
(a) $1.25 cm$
(b) $2.5 cm$
(c) $1.05 cm$
(d) $2 cm$
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Answer:
Correct Answer: 18. (b)
Solution:
- Image formed by convex lens at $I _1$ will act as a virtual object for concave lens. For concave lens
$$ \begin{aligned} \text { or } \quad \frac{1}{v}-\frac{1}{u} & =\frac{1}{f} \\ \text { or } \quad \frac{1}{v}-\frac{1}{4} & =\frac{1}{-20} \\ v & =5 cm \end{aligned} $$
Magnification for concave lens
$$ m=\frac{v}{u}=\frac{5}{4}=1.25 $$
As size of the image at $I _1$ is $2 cm$. Therefore, size of image at $I _2$ will be $2 \times 1.25=2.5 cm$.
