CBSE Class 10 Science 2019 · Paper 31/5/2 · Question 16

Using the lens formula with $f=-60$ cm, $u=-30$ cm and $h=+9$ cm: $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=-\frac{1}{60}-\frac{1}{30}=-\frac{1}{20}$, so $v=-20$ cm. $m=\frac{v}{u}=\frac{-20}{-30}=+\frac{2}{3}$, so $h^{\prime }=m\times h=\frac{2}{3}\times 9=+6$ cm. Nature: virtual, erect and diminished. Position: $20$ cm from the lens on the same side as the object (between the optical centre and the focus $F_1$). Length: $6$ cm. Ray diagram (see the figure): a ray from the top of the object parallel to the axis diverges after the lens as if coming from $F_1$, and a ray through the optical centre goes straight on; the backward extension of the first ray meets the second at the top of the image $A^{\prime }{B}^{\prime }$.