CBSE Class 10 Science 2019 · Paper 31/3/2 · Question 16

(a) When an object is placed between the optical centre and the principal focus of a convex lens, the image formed is: (i) virtual; (ii) erect; (iii) enlarged (magnified); and (iv) on the same side of the lens as the object, farther from the lens than the object. (b) For a concave lens $f=-20$ cm, and its image is virtual and erect, so $m=\frac{v}{u}=+\frac{1}{3}$, which gives $v=\frac{u}{3}$. By the lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$: $\frac{3}{u}-\frac{1}{u}=-\frac{1}{20}$, so $\frac{2}{u}=-\frac{1}{20}$ and $u=-40$ cm. The object is $40$ cm from the lens, on the side from which the light comes (the image is at $v=-\frac{40}{3}\approx -13.3$ cm).