CBSE Class 10 Science 2017 · Paper 31/1/C · Question 9

Ray diagram (drawn to scale for an object $30$ cm from a convex lens of focal length $10$ cm): the object $AB$ stands on the principal axis beyond $2F_1$. A ray from $A$ parallel to the principal axis passes through $F_2$ after refraction, and a ray from $A$ through the optical centre $O$ goes straight on. They meet at $A^{\prime }$ below the axis, so the image $A^{\prime }{B}^{\prime }$ is real and inverted (here diminished, between $F_2$ and $2F_2$). Sign convention: all distances are measured from $O$; those in the direction of the incident light (to the right of the lens) are positive and those against it (to the left) are negative. So the object distance $u=OB$ is negative, the image distance $v=O{B}^{\prime }$ is positive and the focal length $f=O{F}_2$ of the convex lens is positive — in the figure $u=-30$ cm, $v=+15$ cm and $f=+10$ cm. Relation between $u$, $v$ and $f$ (lens formula): $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$.