CBSE Class 10 Science 2018 · Paper 31/1/C · Question 26(b)

Size of the image: $h^{\prime }=-6$ cm, that is, the image is $6$ cm tall and inverted. Working: $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{15}-\frac{1}{25}=\frac{2}{75}$, so $v=+37.5$ cm; $m=\frac{v}{u}=\frac{37.5}{-25}=-1.5$; $h^{\prime }=m\times h=-1.5\times 4=-6$ cm. Ray diagram (the figure at step 4): the object lies between $F_1$ and $2F_1$; the ray parallel to the principal axis is refracted through $F_2$ and the ray through the optical centre O goes on undeviated. They meet $37.5$ cm behind the lens, beyond $2F_2$, where a real, inverted, magnified image $6$ cm tall is formed.