(a) Define optical centre of a spherical lens. (b) A divergent lens has a focal length of 20 cm. At what distance should an object of height 4 cm from the optical centre of the lens be placed so that its image is formed 10 cm away from the lens. Find the size of the image also. (c) Draw a ray diagram to show the formation of image in above situation.

(a) The central point of a spherical lens is called its optical centre. It is usually represented by the letter $O$. A ray of light passing through the optical centre of a lens goes without suffering any deviation.

(b) To find the object distance and image size:

• Given:
• Focal length of divergent (concave) lens, $f = -20\text{ cm}$
• Image distance, $v = -10\text{ cm}$ (since a concave lens always forms a virtual image on the same side as the object)
• Height of the object, $h = 4\text{ cm}$
• Using the lens formula:

$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$

$$\frac{1}{u} = \frac{1}{v} - \frac{1}{f}$$

$$\frac{1}{u} = \frac{1}{-10} - \frac{1}{-20} = -\frac{1}{10} + \frac{1}{20} = \frac{-2 + 1}{20} = -\frac{1}{20}$$

$$u = -20\text{ cm}$$

The object should be placed at a distance of $20\text{ cm}$ in front of the lens.

• Using the magnification formula:

$$m = \frac{v}{u} = \frac{h'}{h}$$

$$h' = h \times \left(\frac{v}{u}\right) = 4\text{ cm} \times \left(\frac{-10\text{ cm}}{-20\text{ cm}}\right) = 4 \times 0.5 = 2\text{ cm}$$

The size of the image is $2\text{ cm}$ (erect and virtual).

(c) Ray Diagram:

An object $AB$ of height $4\text{ cm}$ is placed at $F_1$ ($20\text{ cm}$ from the lens). A ray parallel to the principal axis diverges and appears to pass through $F_1$. Another ray passing through the optical centre $O$ goes undeviated. The virtual, erect, and diminished image $A'B'$ of height $2\text{ cm}$ is formed at $10\text{ cm}$ from the lens on the same side.