CBSE Class 10 Science 2019 · Paper 31/1/1 · Question 19

(i) By the sign convention $u=-60$ cm and $f=-30$ cm. From $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=-\frac{1}{30}-\frac{1}{60}=-\frac{1}{20}$, $v=-20$ cm: the image is 20 cm from the lens, on the same side as the object. The magnification is $m=\frac{v}{u}=\frac{-20}{-60}=+\frac{1}{3}$. (ii) Nature: virtual. Position: 20 cm from the lens on the same side as the object, between O and $F_1$. Size: diminished (one third of the object). Erect. (iii) Ray diagram (see the figure): a ray from the top of the object parallel to the axis spreads out after the lens as if coming from $F_1$, and a ray through the optical centre O goes on undeviated; the backward extension of the first ray meets the second ray 20 cm in front of the lens, giving a virtual, erect and diminished image.