(a) Define focal length of a spherical lens. (b) A divergent lens has a focal length of 30 cm. At what distance should an object of height 5 cm from the optical centre of the lens be placed so that its image is formed 15 cm away from the lens? Find the size of the image also. (c) Draw a ray diagram to show the formation of image in the above situation.

(a) Definition of Focal Length of a Spherical Lens

The distance between the optical centre ($O$) of a spherical lens and its principal focus ($F$) is called its focal length ($f$).

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(b) Numerical Problem

Given:

• Type of lens: Divergent lens (Concave lens)
• Focal length, $f = -30\text{ cm}$ (by sign convention, a concave lens has a negative focal length)
• Image distance, $v = -15\text{ cm}$ (a concave lens always forms a virtual image on the same side as the object)
• Height of the object, $h = 5\text{ cm}$

To find:

• Object distance ($u$)
• Height of the image ($h'$)

Step 1: Find the object distance ($u$) using the Lens Formula

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

$$\frac{1}{-30} = \frac{1}{-15} - \frac{1}{u}$$

$$\frac{1}{u} = -\frac{1}{15} + \frac{1}{30}$$

$$\frac{1}{u} = \frac{-2 + 1}{30} = -\frac{1}{30}$$

$$u = -30\text{ cm}$$

Thus, the object should be placed at a distance of $30\text{ cm}$ in front of the lens.

Step 2: Find the size of the image ($h'$) using the Magnification Formula

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

$$\frac{-15}{-30} = \frac{h'}{5}$$

$$\frac{1}{2} = \frac{h'}{5} \implies h' = 2.5\text{ cm}$$

Thus, the size (height) of the image is $2.5\text{ cm}$. The positive sign indicates that the image is virtual and erect.

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(c) Ray Diagram

Since $u = -30\text{ cm}$ and $f = -30\text{ cm}$, the object is placed exactly at the principal focus ($F_1$) of the concave lens.

``diagram

Concave Lens

Object ||

^ ||

Image
^

----------|-----|-----||--------------------

F1 ||

-30cm -15cm ||

<---v----->
<-------u->

``