CBSE Class 10 Science 2020 · Paper 31/5/1 · Question 30(b)

(a) $f=\frac{1}{P}=\frac{1}{-2.5\text{D}}=-0.4$ m $=-40$ cm; the negative power shows a concave (diverging) lens. A concave lens forms a virtual image on the object's side, so $v=-10$ cm. From $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$: $\frac{1}{u}=\frac{1}{-10}-\frac{1}{-40}=-\frac{3}{40}$, so $u=-\frac{40}{3}$ cm $\approx -13.3$ cm. The object must be placed $\approx 13.3$ cm in front of the lens. (b) Since the power is negative, the lens is concave (diverging). In the ray diagram, a ray from the top of the object parallel to the axis diverges after refraction as if coming from $F_1$ (its backward extension is dashed), and a ray through the optical centre passes undeviated; they meet 10 cm from the lens on the object's side. The image is virtual, erect and diminished ($m=\frac{v}{u}=+0.75$), between the optical centre and the focus $F_1$.