CBSE Class 10 Science 2019 · Paper 31/1/2 · Question 18

(i) With $u=-30$ cm and $f=-30$ cm, $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=-\frac{1}{30}-\frac{1}{30}=-\frac{1}{15}$, so $v=-15$ cm: the image is $15$ cm from the lens, on the same side as the object. (ii) Nature: virtual. Position: $15$ cm from the lens on the same side as the object, between the optical centre and the focus $F_1$. Size: diminished, half the size of the object ($m=v/u=+0.5$). Orientation: erect. (iii) Ray diagram: a ray from the top of the object parallel to the principal axis diverges after refraction so that its backward extension passes through $F_1$, and a ray through the optical centre goes on undeviated. They appear to meet $15$ cm from the lens, which gives a virtual, erect, diminished image, as shown in the figure.