Study the following table for a convex lens for different positions of the object and answer the following questions:
Position of objectPosition of imageRelative size of image
At infinityAt focus $F_2$Highly diminished, point-sized
Beyond $2F_1$Between $F_2$ and $2F_2$Diminished
At $2F_1$At $2F_2$Same size
Between $F_1$ and $2F_1$Beyond $2F_2$Enlarged
At focus $F_1$At infinityInfinitely large or highly enlarged
Between focus $F_1$ and optical centre $O$On the same side of the lens as the objectEnlarged

$(i)$ What is the nature of the image if an object is placed at infinity?
(ii) Identify the nature of the image when the object is between the focus and the optical centre.
(iii) What is the position of the image when the object is placed at focus $F_1$?
(iv) What is the focal length of a lens for an object placed $50 \text{ cm}$ from the lens producing a virtual image at a distance of $10 \text{ cm}$ in front of the lens?

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(N/A) $(i)$ When the object is at infinity, the image is real, inverted, and highly diminished (point-sized).
(ii) When the object is between the focus $F_1$ and the optical centre $O$, the image is virtual and erect.
(iii) When the object is at focus $F_1$, the image is formed at infinity.
(iv) Given:
Object distance $u = -50 \text{ cm}$
Image distance $v = -10 \text{ cm}$ (virtual image is formed on the same side as the object).
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{-10} - \frac{1}{-50} = \frac{1}{f}$
$-\frac{1}{10} + \frac{1}{50} = \frac{1}{f}$
$\frac{-5 + 1}{50} = \frac{1}{f}$
$-\frac{4}{50} = \frac{1}{f}$
$f = -12.5 \text{ cm}$

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MediumCBSE 2014
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