CBSE Class 10 Science 2019 · Paper 31/5/1 · Question 17

(a) Using $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ with $u=-30$ cm and $f=+20$ cm: $\frac{1}{v}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60}$, so $v=+60$ cm. Position: the image is formed $60$ cm from the lens, on the other side of the lens (beyond $2F_2$). Magnification $m=\frac{v}{u}=\frac{60}{-30}=-2$, so $h^{\prime }=m\times h=-2\times 5=-10$ cm. Nature: real, inverted and magnified. Size: $10$ cm tall, twice the size of the object. (b) In the ray diagram, a ray from the top B of the object parallel to the principal axis passes through $F_2$ after refraction, and a ray through the optical centre O goes straight on; they meet at B′, $60$ cm behind the lens, where the real, inverted image A′B′ forms. The object distance $\text{OA}=30$ cm, the image distance ${\text{OA}}^{\prime }=60$ cm and the focal length ${\text{OF}}_2=20$ cm are marked on it.