“A convex lens can form a magnified erect as well as magnified inverted image of an object placed in front of it.” Draw ray diagram to justify this statement stating the position of the object with respect to the lens in each case. An object of height 4 cm is placed at a distance of 20 cm from a concave lens of focal length 10 cm. Use lens formula to determine the position of the image formed.

Part 1: Justification of the Statement

A convex lens can form both a magnified erect image and a magnified inverted image depending on the position of the object relative to its focus ($F_1$) and optical centre ($O$).

#### Case 1: Magnified Erect Image

• Position of Object: Between the principal focus ($F_1$) and the optical centre ($O$) of the lens.
• Nature of Image: Virtual, erect, and magnified.
• Ray Diagram Description:

1. A ray from the top of the object $A$ parallel to the principal axis refracts through the lens and passes through the principal focus $F_2$ on the other side.
2. Another ray from $A$ passes straight through the optical centre $O$ without deviation.
3. These two refracting rays diverge on the right side. When produced backwards, they meet behind the object to form a virtual, erect, and magnified image $A'B'$.

``

Case 1: Object between F1 and O (Virtual, Erect, and Magnified Image)

A'

|\

| \ A

| \ |\

| \ | \

---------B'---|----B--\----(O)------F2------

| \ /

| \/

/

`

#### Case 2: Magnified Inverted Image

• Position of Object: Between $F_1$ and $2F_1$ (or $C_1$) of the lens.
• Nature of Image: Real, inverted, and magnified.
• Ray Diagram Description:

1. A ray from the top of the object $A$ parallel to the principal axis refracts and passes through $F_2$.
2. Another ray from $A$ passes through the optical centre $O$ and continues straight.
3. These two rays intersect on the other side of the lens beyond $2F_2$, forming a real, inverted, and magnified image $A'B'$.

`

Case 2: Object between F1 and 2F1 (Real, Inverted, and Magnified Image)

A

|\

| \

-------2F1----------B--\--F1---(O)---F2--------2F2--------

\ \ |\

\ \ | \

\ \ | \

B' A'

``

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Part 2: Numerical Solution

Given:

• Height of the object, $h = 4\text{ cm}$
• Object distance, $u = -20\text{ cm}$ (by sign convention)
• Focal length of the concave lens, $f = -10\text{ cm}$ (by sign convention)

To find:

• Position of the image, $v$

Using the lens formula:

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

Rearranging the formula to solve for $v$:

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

Substitute the values:

$$\frac{1}{v} = \frac{1}{-10} + \frac{1}{-20}$$

$$\frac{1}{v} = -\frac{1}{10} - \frac{1}{20}$$

$$\frac{1}{v} = \frac{-2 - 1}{20} = -\frac{3}{20}$$

$$v = -\frac{20}{3}\text{ cm} \approx -6.67\text{ cm}$$

Conclusion:

The image is formed at a distance of $6.67\text{ cm}$ from the optical centre of the lens, on the same side as the object. The negative sign indicates that the image is virtual.